message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 1,715 | 14 | 3,430 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
from collections import deque
def bfs(u):
global n, al, dist, prev, used
q = deque()
q.append(u)
md = 1
used[u] = True
while len(q) > 0:
v = q.popleft()
for i in al[v]:
if not used[i]:
used[i] = True
dist[i] = dist[v] + 1
if dist[i] + 1 > md:
md = dist[i] + 1
q.append(i)
return md
n = int(input())
al = [[] for i in range(n)]
dist = [0] * n
prev = [0] * n
used = [False] * n
p = [-1] * n
for i in range(n):
x = int(input()) - 1
if x >= 0:
p[i] = x
al[x].append(i)
res = 0
a = []
for i in range(n):
if p[i] == -1:
a.append(i)
i = 0
while i < len(a):
if not used[a[i]]:
res = max(res, bfs(a[i]))
else:
i += 1
print(res)
``` | output | 1 | 1,715 | 14 | 3,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 1,716 | 14 | 3,432 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
def findDepth(a, i):
depth = 1
nextLevel = a[i][:]
while len(nextLevel) > 0:
depth += 1
children = nextLevel[:]
nextLevel = []
for child in children:
nextLevel += a[child]
return depth
n = int(input())
a = []
for i in range(n):
a.append([])
roots = []
for i in range(n):
x = int(input())
if x > 0:
a[x-1].append(i)
else:
roots.append(i)
print(max([findDepth(a, i) for i in roots]))
``` | output | 1 | 1,716 | 14 | 3,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5 | instruction | 0 | 1,717 | 14 | 3,434 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
n=int(input())
p=[int(input())-1 for i in range(n)]
m=0
for i in range(len(p)):
c=0
while i!=-2:
c+=1
i=p[i]
m=max(c,m)
print(m)
# Made By Mostafa_Khaled
``` | output | 1 | 1,717 | 14 | 3,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
from sys import setrecursionlimit
setrecursionlimit(30000)
maxl = -1
def dfs(v, lvl):
used[v] = True
global maxl
if lvl > maxl:
maxl = lvl
for u in ch[v]:
dfs(u, lvl + 1)
n = int(input())
used = [False] * n
p = [int(input()) - 1 for i in range(n)]
ch = [[] for i in range(n)]
for i in range(n):
if p[i] != -2:
ch[p[i]].append(i)
for i in range(n):
if p[i] == -2:
dfs(i, 1)
print(maxl)
``` | instruction | 0 | 1,718 | 14 | 3,436 |
Yes | output | 1 | 1,718 | 14 | 3,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
from collections import *
class graph:
# initialize graph
def __init__(self, gdict=None):
if gdict is None:
gdict = defaultdict(list)
self.gdict, self.edges = gdict, []
# Get verticies
def get_vertices(self):
return list(self.gdict.keys())
# add edge
def add_edge(self, node1, node2):
self.gdict[node1].append(node2)
# self.gdict[node2].append(node1)
self.edges.append([node1, node2])
def bfs_util(self, i):
queue, self.visit[i], level = deque([[i, 1]]), 1, 1
while queue:
# dequeue parent vertix
s, level = queue.popleft()
# enqueue child vertices
for i in self.gdict[s]:
if self.visit[i] == 0:
queue.append([i, level + 1])
self.visit[i] = 1
return level
def bfs(self):
self.visit, self.cnt = defaultdict(int), 0
for i in pars:
if self.visit[i] == 0:
self.cnt = max(self.bfs_util(i), self.cnt)
return self.cnt
n, g, pars = int(input()), graph(), []
for i in range(1, n + 1):
x = int(input())
if x == -1:
pars.append(i)
else:
g.add_edge(x, i)
print(g.bfs())
``` | instruction | 0 | 1,719 | 14 | 3,438 |
Yes | output | 1 | 1,719 | 14 | 3,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
inp,m = [int(input())-1 for n in range(int(input()))],0
for i in range(len(inp)):
c = 0
while i != -2:
c+=1; i = inp[i]
m = max(c,m)
print(m)
``` | instruction | 0 | 1,720 | 14 | 3,440 |
Yes | output | 1 | 1,720 | 14 | 3,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
# Hint: Find the longest chain of command.
size =int(input())
lis = [-1]*(size)
for y in range(size):
lis[y] = int(input()) - 1
r = [0]*(size)
for x in range(size):
c = x
while lis[c] >-1:
# Adding the employee to the manager
r[x] += 1
# Moving up the chain
c = lis[c]
print (max(r)+1)
``` | instruction | 0 | 1,721 | 14 | 3,442 |
Yes | output | 1 | 1,721 | 14 | 3,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
from collections import deque
def bfs(u):
global n, al, dist, prev, used
q = deque()
q.append(u)
md = 1
used[u] = True
while len(q) > 0:
v = q.popleft()
for i in al[v]:
if not used[i]:
used[i] = True
dist[i] = dist[v] + 1
if dist[i] + 1 > md:
md = dist[i] + 1
q.append(i)
return md
n = int(input())
al = [[] for i in range(n)]
dist = [0] * n
prev = [0] * n
used = [False] * n
for i in range(n):
x = int(input()) - 1
if x >= 0:
al[x].append(i)
res = 0
for i in range(n):
if not used[i]:
res = max(res, bfs(i))
print(res)
``` | instruction | 0 | 1,722 | 14 | 3,444 |
No | output | 1 | 1,722 | 14 | 3,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
n=int(input())
a=[]
for i in range(n+1):
a.append([])
for i in range(1,n+1):
p=int(input())
if p!=-1:
a[p].append(i)
ans=1
for b in a:
if len(b)>0: ans=2
for x in b:
if len(a[x])>0:
print(3)
exit()
print(ans)
``` | instruction | 0 | 1,723 | 14 | 3,446 |
No | output | 1 | 1,723 | 14 | 3,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():
return map(int,input().split())
def array():
return [int(i) for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
def bfs(adj,s,n,vis):
vis[s]=True
qu=[s]
c=0
while(len(qu)!=0):
e=qu.pop(0)
flag=0
for i in adj[e]:
if(vis[i]!=True):
flag=1
qu.append(i)
vis[i]=True
if(flag==1):
c+=1
return c+1
adj=defaultdict(list)
n=int(input())
for i in range(n):
a=int(input())
if(a!=-1):
adj[a].append(i+1)
else:
adj[i+1]=[]
#print(adj)
ans=0
vis={i:False for i in range(1,n+1)}
for i in range(1,n+1):
if(vis[i]==False):
ans=max(ans,bfs(adj,i,n,vis))
#print(bfs(adj,i,n,vis),i)
print(ans)
``` | instruction | 0 | 1,724 | 14 | 3,448 |
No | output | 1 | 1,724 | 14 | 3,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:
* Employee A is the immediate manager of employee B
* Employee B has an immediate manager employee C such that employee A is the superior of employee C.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.
What is the minimum number of groups that must be formed?
Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.
The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.
Output
Print a single integer denoting the minimum number of groups that will be formed in the party.
Examples
Input
5
-1
1
2
1
-1
Output
3
Note
For the first example, three groups are sufficient, for example:
* Employee 1
* Employees 2 and 4
* Employees 3 and 5
Submitted Solution:
```
n=int(input())
a=[]
for i in range(n+1):
a.append([])
for i in range(1,n+1):
p=int(input())
if p!=-1:
a[p].append(i)
print(a)
ans=1
for b in a:
if len(b)>0: ans=2
for x in b:
if len(a[x])>0:
print(3)
exit()
print(ans)
``` | instruction | 0 | 1,725 | 14 | 3,450 |
No | output | 1 | 1,725 | 14 | 3,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
def main():
n,k = map(int, input().split());pp = [int(a)-1 for a in input().split()];chainhead = [True]*n;follower = [None]*n
for _ in range(k):
x, y = map(int, input().split())
if follower[x-1]:print(0);return
else:follower[x-1] = y-1;chainhead[y-1] = False
chain = list(range(n));cd = [0]*n
for i in range(n):
if chainhead[i]:
f = follower[i];d = 1
while f != None:cd[f] = d;d += 1;chain[f] = i;f = follower[f]
chainparents = [[] for _ in range(n)];ccount = [0]*n
for i in range(n):
if pp[i] != -1:
if chain[i] != chain[pp[i]]:chainparents[chain[i]].append(chain[pp[i]]);ccount[chain[pp[i]]] += 1
elif cd[pp[i]] > cd[i]:print(0);return
s = [i for i in range(n) if chainhead[i] and ccount[i] == 0];l = [];res = []
while s:
v = s.pop();l.append(v)
for p in chainparents[v]:
ccount[p] -= 1
if ccount[p] == 0:s.append(p)
if any(ccount[i] != 0 for i in range(n)):print(0);return
for h in l[::-1]:
c = h
while c != None:res.append(c+1);c = follower[c]
print(' '.join(map(str, res)))
main()
``` | instruction | 0 | 1,861 | 14 | 3,722 |
Yes | output | 1 | 1,861 | 14 | 3,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
class Node:
def __init__(self, value):
self.value = value
self.next = None
self.prev = None
def __repr__(self):
if not self:
return "{}()".format(self.__class__.__name__)
return "{}({})".format(self.__class__.__name__, self.value)
class LL:
def __init__(self, iterable=None):
self.sentinel = Node(None)
self.sentinel.next = self.sentinel
self.sentinel.prev = self.sentinel
self.__len = 0
if iterable is not None:
self += iterable
def get_node(self, index):
node = sentinel = self.sentinel
i = 0
while i <= index:
node = node.next
if node == sentinel:
break
i += 1
if node == sentinel:
node = None
return node
def __getitem__(self, index):
node = self.get_node(index)
return node.value
def __len__(self):
return self.__len
def __setitem__(self, index, value):
node = self.get_node(index)
node.value = value
def __delitem__(self, index):
node = self.get_node(index)
if node:
node.prev.next = node.next
if node.next:
node.next.prev = node.prev
node.prev = None
node.next = None
node.value = None
self.__len -= 1
def __repr__(self):
return str(self.to_list())
def to_list(self):
l = []
c = self.sentinel.next
while c != self.sentinel:
l.append(c.value)
c = c.next
return l
def append(self, value):
sentinel = self.sentinel
node = Node(value)
self.insert_between(node, sentinel.prev, sentinel)
def appendleft(self, value):
sentinel = self.sentinel
node = Node(value)
self.insert_between(node, sentinel, sentinel.next)
def insert(self, index, value):
sentinel = self.sentinel
new_node = Node(value)
len_ = len(self)
if len_ == 0:
self.insert_between(new_node, sentinel, sentinel)
elif index >= 0 and index < len_:
node = self.get_node(index)
self.insert_between(new_node, node.prev, node)
elif index == len_:
self.insert_between(new_node, sentinel.prev, sentinel)
else:
raise IndexError
self.__len += 1
def insert_between(self, node, left_node, right_node):
if node and left_node and right_node:
node.prev = left_node
node.next = right_node
left_node.next = node
right_node.prev = node
else:
raise IndexError
def merge_left(self, other):
sentinel = self.sentinel
sentinel.next.prev = other.sentinel.prev
other.sentinel.prev.next = sentinel.next
sentinel.next = other.sentinel.next
sentinel.next.prev = sentinel
def merge_right(self, other):
sentinel = self.sentinel
sentinel.prev.next = other.sentinel.next
other.sentinel.next.prev = sentinel.prev
sentinel.prev = other.sentinel.prev
sentinel.prev.next = sentinel
import sys,io,os#;Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
Z=sys.stdin.readline
X=lambda:print(0)or quit()
from collections import deque
def path(R):
H=deque();H.append(R)
while P[R]>=0:
R=P[R];H.append(R)
if len(H)>2:P[H.popleft()]=H[-1]
return R
N,k=map(int,Z().split());p=[*map(int,Z().split())]
K=[-1]*N;P=[-1]*(N//2);S=[1]*(N//2);ch=[];R=0
for _ in range(k):
a,b=map(int,Z().split());a-=1;b-=1
if K[a]>=0:
if K[b]>=0:
va,vb=path(K[a]),path(K[b])
if va!=vb:
sa,sb=S[va],S[vb]
if sa>sb:
P[vb]=va;ch[va].merge_right(ch[vb]);ch[vb]=0
else:
P[va]=vb;ch[vb].merge_left(ch[va]);ch[va]=0
if sa==sb:S[vb]+=1
else:X()
else:va=path(K[a]);K[b]=va;ch[va].append(b)
else:
if K[b]>=0:vb=path(K[b]);K[a]=vb;ch[vb].appendleft(a)
else:
K[a]=R;K[b]=R;R+=1
l=LL();l.append(a);l.append(b)
ch.append(l)
f=[[]for i in range(N)];x=[-1]*N;h=[0]*N;u=set()
for z in ch:
if z!=0:
c=z.sentinel.next;i=0
while c!=z.sentinel:x[c.value]=i;i+=1;c=c.next
for i in range(N):
a,b=p[i]-1,i;va,vb=a,b
if a<0:z=b;continue
if K[a]>=0:va=ch[path(K[a])][0]
if K[b]>=0:vb=ch[path(K[b])][0]
if va==vb:
if x[a]>x[b]:X()
else:
if(va,vb)not in u:f[va].append(vb);h[vb]+=1;u.add((va,vb))
q=[z];o=[];u=set()
while q:
v=q.pop()
if v in u:X()
u.add(v)
if x[v]>=0:o+=ch[path(K[v])].to_list()
else:o.append(v)
for i in f[v]:
h[i]-=1
if h[i]<1:q.append(i)
if len(o)<N:X()
print(' '.join(map(lambda i:str(i+1),o)))
``` | instruction | 0 | 1,862 | 14 | 3,724 |
Yes | output | 1 | 1,862 | 14 | 3,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
class Node:
def __init__(self, value):
self.value = value
self.next = None
self.prev = None
def __repr__(self):
if not self:
return "{}()".format(self.__class__.__name__)
return "{}({})".format(self.__class__.__name__, self.value)
class LL:
def __init__(self, iterable=None):
self.sentinel = Node(None)
self.sentinel.next = self.sentinel
self.sentinel.prev = self.sentinel
self.__len = 0
if iterable is not None:
self += iterable
def get_node(self, index):
node = sentinel = self.sentinel
i = 0
while i <= index:
node = node.next
if node == sentinel:
break
i += 1
if node == sentinel:
node = None
return node
def __getitem__(self, index):
node = self.get_node(index)
return node.value
def __len__(self):
return self.__len
def __setitem__(self, index, value):
node = self.get_node(index)
node.value = value
def __delitem__(self, index):
node = self.get_node(index)
if node:
node.prev.next = node.next
if node.next:
node.next.prev = node.prev
node.prev = None
node.next = None
node.value = None
self.__len -= 1
def __repr__(self):
return str(self.to_list())
def to_list(self):
l = []
c = self.sentinel.next
while c != self.sentinel:
l.append(c.value)
c = c.next
return l
def append(self, value):
sentinel = self.sentinel
node = Node(value)
self.insert_between(node, sentinel.prev, sentinel)
def appendleft(self, value):
sentinel = self.sentinel
node = Node(value)
self.insert_between(node, sentinel, sentinel.next)
def insert(self, index, value):
sentinel = self.sentinel
new_node = Node(value)
len_ = len(self)
if len_ == 0:
self.insert_between(new_node, sentinel, sentinel)
elif index >= 0 and index < len_:
node = self.get_node(index)
self.insert_between(new_node, node.prev, node)
elif index == len_:
self.insert_between(new_node, sentinel.prev, sentinel)
else:
raise IndexError
self.__len += 1
def insert_between(self, node, left_node, right_node):
if node and left_node and right_node:
node.prev = left_node
node.next = right_node
left_node.next = node
right_node.prev = node
else:
raise IndexError
def merge_left(self, other):
sentinel = self.sentinel
sentinel.next.prev = other.sentinel.prev
other.sentinel.prev.next = sentinel.next
sentinel.next = other.sentinel.next
sentinel.next.prev = sentinel
def merge_right(self, other):
sentinel = self.sentinel
sentinel.prev.next = other.sentinel.next
other.sentinel.next.prev = sentinel.prev
sentinel.prev = other.sentinel.prev
sentinel.prev.next = sentinel
import sys,io,os;Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
X=lambda:print(0)or quit()
from collections import deque
def path(R):
H=deque();H.append(R)
while P[R]>=0:
R=P[R];H.append(R)
if len(H)>2:P[H.popleft()]=H[-1]
return R
N,k=map(int,Z().split());p=[*map(int,Z().split())]
K=[-1]*N;P=[-1]*(N//2);S=[1]*(N//2);ch=[];R=0
for _ in range(k):
a,b=map(int,Z().split());a-=1;b-=1
if K[a]>=0:
if K[b]>=0:
va,vb=path(K[a]),path(K[b])
if va!=vb:
sa,sb=S[va],S[vb]
if sa>sb:
P[vb]=va;ch[va].merge_right(ch[vb]);ch[vb]=0
else:
P[va]=vb;ch[vb].merge_left(ch[va]);ch[va]=0
if sa==sb:S[vb]+=1
else:X()
else:va=path(K[a]);K[b]=va;ch[va].append(b)
else:
if K[b]>=0:vb=path(K[b]);K[a]=vb;ch[vb].appendleft(a)
else:
K[a]=R;K[b]=R;R+=1
l=LL();l.append(a);l.append(b)
ch.append(l)
f=[[]for i in range(N)];x=[-1]*N;h=[0]*N;u=set()
for z in ch:
if z!=0:
c=z.sentinel.next;i=0
while c!=z.sentinel:x[c.value]=i;i+=1;c=c.next
for i in range(N):
a,b=p[i]-1,i;va,vb=a,b
if a<0:z=b;continue
if K[a]>=0:va=ch[path(K[a])][0]
if K[b]>=0:vb=ch[path(K[b])][0]
if va==vb:
if x[a]>x[b]:X()
else:
if(va,vb)not in u:f[va].append(vb);h[vb]+=1;u.add((va,vb))
q=[z];o=[];u=set()
while q:
v=q.pop()
if v in u:X()
u.add(v)
if x[v]>=0:o+=ch[path(K[v])].to_list()
else:o.append(v)
for i in f[v]:
h[i]-=1
if h[i]<1:q.append(i)
if len(o)<N:X()
print(' '.join(map(lambda i:str(i+1),o)))
``` | instruction | 0 | 1,863 | 14 | 3,726 |
Yes | output | 1 | 1,863 | 14 | 3,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
n,k=map(int,input().split())
P=list(map(int,input().split()))
T=[tuple(map(int,input().split())) for i in range(k)]
for i in range(n):
if P[i]==0:
initial=i+1
D=dict()
D2=dict()
# UnionFind
Group = [i for i in range(n+1)]
Nodes = [1]*(n+1)
def find(x):
while Group[x] != x:
x=Group[x]
return x
def Union(x,y):
if find(x) != find(y):
if Nodes[find(x)] < Nodes[find(y)]:
Nodes[find(y)] += Nodes[find(x)]
Nodes[find(x)] = 0
Group[find(x)] = find(y)
else:
Nodes[find(x)] += Nodes[find(y)]
Nodes[find(y)] = 0
Group[find(y)] = find(x)
for x,y in T:
Union(x,y)
if x in D:
print(0)
sys.exit()
if y in D2:
print(0)
sys.exit()
if y==initial:
print(0)
sys.exit()
D[x]=y
D2[y]=x
E=[set() for i in range(n+1)]
for i in range(n):
if P[i]==0:
initial=i+1
elif P[i] in D2 and D2[P[i]]==i+1:
print(0)
sys.exit()
else:
E[find(P[i])].add(find(i+1))
EDGEIN=[0]*(n+1)
for i in range(n+1):
for to in E[i]:
if find(i)==find(to):
continue
EDGEIN[to]+=1
Q=[]
for i in range(1,n+1):
if find(i)!=i:
continue
if EDGEIN[i]==0:
Q.append(i)
ANS=[]
USE=[0]*(n+1)
while Q:
#print(Q,ANS)
x=Q.pop()
if USE[x]==1:
print(0)
sys.exit()
else:
#ANS.append(x)
USE[x]=0
z=x
count=0
while z in D2:
z=D2[z]
count+=1
if count>n+5:
print(0)
sys.exit()
ANS.append(z)
while z in D:
if USE[D[z]]==0:
ANS.append(D[z])
z=D[z]
USE[z]=1
else:
print(0)
sys.exit()
for to in E[x]:
EDGEIN[to]-=1
if EDGEIN[to]==0:
Q.append(to)
if len(ANS)==n:
print(*ANS)
else:
print(0)
``` | instruction | 0 | 1,864 | 14 | 3,728 |
Yes | output | 1 | 1,864 | 14 | 3,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
n,k=map(int,input().split())
P=list(map(int,input().split()))
T=[tuple(map(int,input().split())) for i in range(k)]
E=[[] for i in range(n+1)]
EFROM=[[] for i in range(n+1)]
for i in range(n):
if P[i]==0:
initial=i+1
else:
E[P[i]].append(i+1)
EFROM[i+1].append(P[i])
D=dict()
D2=dict()
for x,y in T:
if x in D:
print(0)
sys.exit()
if y in D2:
print(0)
sys.exit()
if y==initial:
print(0)
sys.exit()
D[x]=y
D2[y]=x
EDGEOUT=[0]*(n+1)
for i in range(n+1):
for to in E[i]:
EDGEOUT[i]+=1
for x,y in T:
EDGEOUT[x]+=1
Q=[]
Q2=[]
QSOON=-1
for i in range(n+1):
if EDGEOUT[i]==0 and i!=0:
if i in D:
continue
if i in D2:
Q2.append(i)
else:
Q.append(i)
ANS=[]
USE=[0]*(n+1)
while Q or Q2 or QSOON!=-1:
#print(Q,Q2,QSOON)
if QSOON!=-1:
x=QSOON
QSOON=-1
if USE[x]==1:
print(0)
sys.exit()
USE[x]=1
ANS.append(x)
fr=P[x-1]
EDGEOUT[fr]-=1
if x in D2:
QSOON=D2[x]
EDGEOUT[D2[x]]-=1
if D2[x]==fr:
continue
else:
if EDGEOUT[fr]==0:
if fr in D2:
Q2.append(fr)
else:
Q.append(fr)
else:
if EDGEOUT[fr]==0:
if fr in D2:
Q2.append(fr)
else:
Q.append(fr)
elif Q:
x=Q.pop()
if USE[x]==1:
continue
USE[x]=1
ANS.append(x)
fr=P[x-1]
EDGEOUT[fr]-=1
if EDGEOUT[fr]==0:
if fr in D2:
Q2.append(fr)
else:
Q.append(fr)
else:
x=Q2.pop()
if USE[x]==1:
continue
USE[x]=1
ANS.append(x)
fr=P[x-1]
EDGEOUT[fr]-=1
QSOON=D2[x]
EDGEOUT[D2[x]]-=1
if D2[x]==fr:
continue
else:
if EDGEOUT[fr]==0:
if fr in D2:
Q2.append(fr)
else:
Q.append(fr)
ANS.reverse()
if len(ANS)==n:
USE=[0]*(n+1)
for i in range(n-1,-1,-1):
if USE[P[ANS[i]-1]]==1:
print(0)
sys.exit()
USE[ANS[i]]=1
print(*ANS)
else:
print(0)
``` | instruction | 0 | 1,865 | 14 | 3,730 |
No | output | 1 | 1,865 | 14 | 3,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
import io
import os
from collections import defaultdict
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def toposort(g):
inc = {}
for x in g:
inc[x] = 0
for l in g.values():
for y in l:
inc[y] += 1
no_inc = [x for x in g if inc[x] == 0]
ans = []
found = 0
while no_inc:
x = no_inc.pop()
ans.append(x)
found += 1
for nei in g[x]:
inc[nei] -= 1
if inc[nei] == 0:
no_inc.append(nei)
if found < len(g):
return None
return ans
def solve():
n, k = map(int, input().split())
p = list(map(int, input().split()))
edges = []
for _ in range(k):
x, y = map(int, input().split())
x -= 1
y -= 1
edges.append((x, y))
g = {i: set() for i in range(n)}
for x, y in edges:
g[x].add(y)
for i, v in enumerate(p):
if v != 0:
g[v-1].add(i)
ans = toposort(g)
if ans is None:
print(0)
return
next = [-1] * n
inc = [0] * n
for x, y in edges:
next[x] = y
inc[y] += 1
no_inc = [v for v in range(n) if inc[v] == 0]
rep = list(range(n))
while no_inc:
v = no_inc.pop()
if next[v] != -1:
rep[next[v]] = rep[v]
no_inc.append(next[v])
g = {x: set() for x in set(rep)}
for i, v in enumerate(p):
if v != 0:
if rep[v-1] != rep[i]:
g[rep[v-1]].add(rep[i])
ans = toposort(g)
if ans is None:
print(0, 1)
return
final_ans = []
for x in ans:
final_ans.append(x)
while next[x] != -1:
x = next[x]
final_ans.append(x)
print(*[x+1 for x in final_ans])
t = 1
for _ in range(t):
solve()
``` | instruction | 0 | 1,866 | 14 | 3,732 |
No | output | 1 | 1,866 | 14 | 3,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
import io
import os
from collections import defaultdict
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def toposort(g):
inc = {}
for x in g:
inc[x] = 0
for l in g.values():
for y in l:
inc[y] += 1
no_inc = [x for x in g if inc[x] == 0]
ans = []
found = 0
while no_inc:
x = no_inc.pop()
ans.append(x)
found += 1
for nei in g[x]:
inc[nei] -= 1
if inc[nei] == 0:
no_inc.append(nei)
if found < len(g):
return None
return ans
def solve():
n, k = map(int, input().split())
p = list(map(int, input().split()))
edges = []
for _ in range(k):
x, y = map(int, input().split())
x -= 1
y -= 1
edges.append((x, y))
g = {i: set() for i in range(n)}
for x, y in edges:
g[x].add(y)
for i, v in enumerate(p):
if v != 0:
g[v-1].add(i)
ans = toposort(g)
if ans is None:
print(0)
return
next = [-1] * n
inc = [0] * n
for i, v in enumerate(p):
if v != 0:
g[v-1] = i
inc[i] += 1
no_inc = [v for v in range(n) if inc[v] == 0]
rep = list(range(n))
while no_inc:
v = no_inc.pop()
if next[v] != -1:
rep[next[v]] = rep[v]
g = {x: set() for x in set(rep)}
for x, y in edges:
g[rep[x]].add(rep[y])
ans = toposort(g)
if ans is None:
print(0)
return
final_ans = []
for x in ans:
final_ans.append(x)
while next[x] != -1:
x = next[x]
final_ans.append(x)
print(*[x+1 for x in final_ans])
t = 1
for _ in range(t):
solve()
``` | instruction | 0 | 1,867 | 14 | 3,734 |
No | output | 1 | 1,867 | 14 | 3,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n (let's call this permutation q). q_i is the index of the topic Ivan will explain during the i-th lecture.
For each topic (except exactly one), there exists a prerequisite topic (for the topic i, the prerequisite topic is p_i). Ivan cannot give a lecture on a topic before giving a lecture on its prerequisite topic. There exists at least one valid ordering of topics according to these prerequisite constraints.
Ordering the topics correctly can help students understand the lectures better. Ivan has k special pairs of topics (x_i, y_i) such that he knows that the students will understand the y_i-th topic better if the lecture on it is conducted right after the lecture on the x_i-th topic. Ivan wants to satisfy the constraints on every such pair, that is, for every i ∈ [1, k], there should exist some j ∈ [1, n - 1] such that q_j = x_i and q_{j + 1} = y_i.
Now Ivan wants to know if there exists an ordering of topics that satisfies all these constraints, and if at least one exists, find any of them.
Input
The first line contains two integers n and k (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ n - 1) — the number of topics and the number of special pairs of topics, respectively.
The second line contains n integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ n), where p_i is the prerequisite topic for the topic i (or p_i = 0 if the i-th topic has no prerequisite topics). Exactly one of these integers is 0. At least one ordering of topics such that for every i the p_i-th topic is placed before the i-th topic exists.
Then k lines follow, the i-th line contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the topics from the i-th special pair. All values of x_i are pairwise distinct; similarly, all valus of y_i are pairwise distinct.
Output
If there is no ordering of topics meeting all the constraints, print 0.
Otherwise, print n pairwise distinct integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — the ordering of topics meeting all of the constraints. If there are multiple answers, print any of them.
Examples
Input
5 2
2 3 0 5 3
1 5
5 4
Output
3 2 1 5 4
Input
5 2
2 3 0 5 3
1 5
5 1
Output
0
Input
5 1
2 3 0 5 3
4 5
Output
0
Input
5 4
2 3 0 5 3
2 1
3 5
5 2
1 4
Output
3 5 2 1 4
Submitted Solution:
```
import sys
input = sys.stdin.readline
from collections import deque
n,k=map(int,input().split())
P=list(map(int,input().split()))
T=[tuple(map(int,input().split())) for i in range(k)]
E=[[] for i in range(n+1)]
for i in range(n):
if P[i]==0:
initial=i+1
else:
E[P[i]].append(i+1)
D=dict()
D2=dict()
for x,y in T:
if x in D:
print(0)
sys.exit()
if y in D2:
print(0)
sys.exit()
D[x]=y
D2[y]=x
ANS=[]
Q=deque([initial])
while Q:
x=Q.popleft()
ANS.append(x)
if x in D:
Q.appendleft(D[x])
continue
for to in E[x]:
if to in D2:
continue
else:
Q.append(to)
if len(ANS)==n:
print(*ANS)
else:
print(0)
``` | instruction | 0 | 1,868 | 14 | 3,736 |
No | output | 1 | 1,868 | 14 | 3,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,869 | 14 | 3,738 |
Tags: dp
Correct Solution:
```
f = lambda: map(int, input().split())
n, m = f()
u = [0]
for i in range(n):
t = list(f())
k = t[0]
for i in range(1, k): t[i + 1] += t[i]
v = [t[k]] * (k + 1)
v[0] = t[0] = 0
for d in range(1, k):
v[k - d] -= min(t[j + d] - t[j] for j in range(k - d + 1))
p = [0] * (min(m, len(u) + len(v)) + 1)
for i, x in enumerate(u):
for j, y in enumerate(v, i):
if j > m: break
p[j] = max(p[j], x + y)
u = p
print(u[m])
``` | output | 1 | 1,869 | 14 | 3,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,870 | 14 | 3,740 |
Tags: dp
Correct Solution:
```
def mem(ind,j):
if(ind == n):
return 0
if(dp2[ind][j] != -1):
return dp2[ind][j]
ans = 0
for i in range(min(j,sz[ind])+1):
ans = max(ans,dp[ind][i]+mem(ind+1,j-i))
dp2[ind][j] = ans
return ans
n,m = map(int,input().split())
sz = [0 for i in range(n)]
dp = []
for y in range(n):
a = list(map(int,input().split()))
sz[y] = a[0]
pre = [a[1]]
for i in range(2,sz[y]+1):
pre.append(a[i]+pre[-1])
s = sum(a[1:])
dp1 = [0 for j in range(sz[y]+1)]
dp1[sz[y]] = s
for i in range(sz[y]):
for j in range(i,sz[y]):
if(i != 0): dp1[sz[y]-(j-i+1)] = max(dp1[sz[y]-(j-i+1)],s-pre[j]+pre[i-1])
else: dp1[sz[y]-(j-i+1)] = max(dp1[sz[y]-(j-i+1)],s-pre[j])
dp.append(dp1)
dp2 = [[-1 for i in range(m+1)] for j in range(n)]
print(mem(0,m))
``` | output | 1 | 1,870 | 14 | 3,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,871 | 14 | 3,742 |
Tags: dp
Correct Solution:
```
f = lambda: list(map(int, input().split()))
def g(t):
n = t[0]
for i in range(1, n): t[i + 1] += t[i]
p = [t[n]] * (n + 1)
p[n] = t[0] = 0
for d in range(1, n):
p[d] -= min(t[j + d] - t[j] for j in range(n - d + 1))
return p[::-1]
n, m = f()
u = g(f())
for i in range(n - 1):
v = g(f())
p = [0] * (min(m, len(u) + len(v)) + 1)
for i, x in enumerate(u):
for j, y in enumerate(v, i):
if j > m: break
p[j] = max(p[j], x + y)
u = p
print(u[m])
``` | output | 1 | 1,871 | 14 | 3,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,872 | 14 | 3,744 |
Tags: dp
Correct Solution:
```
import io, os
input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
n, m = map(int, input().split())
dp = [0] * (m + 1)
for i in range(n):
a = list(map(int, input().split()))
t = sum(a[1:])
b = [0] * (a[0] + 1)
b[-1] = t
for i in range(0, len(a)):
s = t
for j in range(i + 1, len(a)):
s -= a[j]
b[a[0] - j + i] = max(b[a[0] - j + i], s)
for x in range(m, -1, -1):
for y in range(1, min(x, a[0]) + 1):
dp[x] = max(dp[x], dp[x - y] + b[y])
print(dp[m])
``` | output | 1 | 1,872 | 14 | 3,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,873 | 14 | 3,746 |
Tags: dp
Correct Solution:
```
f = lambda: list(map(int, input().split()))
def g(t):
n = t[0]
t[0] = 0
for i in range(1, n): t[i + 1] += t[i]
p = [t[n]] * (n + 1)
p[n] = 0
for d in range(1, n):
p[d] -= min(t[j + d] - t[j] for j in range(n - d + 1))
p.reverse()
return p
def h(u):
v = g(f())
p = [0] * (min(m, len(u) + len(v)) + 1)
for i, x in enumerate(u):
for j, y in enumerate(v):
if i + j > m: break
p[i + j] = max(p[i + j], x + y)
return p
n, m = f()
u = g(f())
for i in range(n - 1): u = h(u)
print(u[m])
``` | output | 1 | 1,873 | 14 | 3,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,874 | 14 | 3,748 |
Tags: dp
Correct Solution:
```
#!/usr/bin/env python3
import os
import sys
from io import BytesIO, IOBase
class FastO:
def __init__(self, fd=1):
stream = BytesIO()
self.flush = lambda: os.write(fd, stream.getvalue()) and not stream.truncate(0) and stream.seek(0)
self.write = lambda b: stream.write(b.encode())
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
sys.stdout, cout = FastO(), ostream()
numbers, num, sign = [], 0, True
for char in os.read(0, os.fstat(0).st_size):
if char >= 48:
num = num * 10 + char - 48
elif char == 45:
sign = False
elif char != 13:
numbers.append(num if sign else -num)
num, sign = 0, True
if char >= 48:
numbers.append(num if sign else -num)
getnum = iter(numbers).__next__
def main():
n, m = getnum(), getnum()
_dp = [[] for _ in range(n)]
for i in range(n):
ni = getnum()
ai = [getnum() for _ in range(ni)]
cumsum = [0] * (ni + 1)
for j in range(ni):
cumsum[j + 1] = cumsum[j] + ai[j]
_dpi = [0] * (ni + 1)
for j in range(ni + 1):
for k in range(j, ni + 1):
_dpi[ni + j - k] = max(_dpi[ni + j - k], cumsum[j] + cumsum[-1] - cumsum[k])
_dp[i] = _dpi
dp = [0] * (m + 1)
for i in range(n):
for j in reversed(range(m + 1)):
for k in range(min(len(_dp[i]), j + 1)):
dp[j] = max(dp[j], _dp[i][k] + dp[j - k])
cout << dp[m]
if __name__ == '__main__':
main()
``` | output | 1 | 1,874 | 14 | 3,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,875 | 14 | 3,750 |
Tags: dp
Correct Solution:
```
#!/usr/bin/env python3
import os
import sys
from io import BytesIO, IOBase
class FastO:
def __init__(self, fd=1):
stream = BytesIO()
self.flush = lambda: os.write(fd, stream.getvalue()) and not stream.truncate(0) and stream.seek(0)
self.write = lambda b: stream.write(b.encode())
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
sys.stdout, cout = FastO(), ostream()
numbers, num, sign = [], 0, True
for char in os.read(0, os.fstat(0).st_size):
if char >= 48:
num = num * 10 + char - 48
elif char == 45:
sign = False
elif char != 13:
numbers.append(num if sign else -num)
num, sign = 0, True
if char >= 48:
numbers.append(num if sign else -num)
getnum = numbers[::-1].pop
def main():
n, m = getnum(), getnum()
_dp = [[] for _ in range(n)]
for i in range(n):
ni = getnum()
ai = [getnum() for _ in range(ni)]
cumsum = [0] * (ni + 1)
for j in range(ni):
cumsum[j + 1] = cumsum[j] + ai[j]
_dpi = [0] * (ni + 1)
for j in range(ni + 1):
for k in range(j, ni + 1):
_dpi[ni + j - k] = max(_dpi[ni + j - k], cumsum[j] + cumsum[-1] - cumsum[k])
_dp[i] = _dpi
dp = [0] * (m + 1)
for i in range(n):
_dpi = _dp[i]
for j in reversed(range(m + 1)):
dpj = dp[j]
for k in range(min(len(_dp[i]), j + 1)):
dpj = max(dpj, _dpi[k] + dp[j - k])
dp[j] = dpj
cout << dp[m]
if __name__ == '__main__':
main()
``` | output | 1 | 1,875 | 14 | 3,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.
The collection of porcelain is arranged neatly on n shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.
You are given the values of all items. Your task is to find the maximal damage the princess' tantrum of m shrieks can inflict on the collection of porcelain.
Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at least m.
Output
Output the maximal total value of a tantrum of m shrieks.
Examples
Input
2 3
3 3 7 2
3 4 1 5
Output
15
Input
1 3
4 4 3 1 2
Output
9
Note
In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.
In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side. | instruction | 0 | 1,876 | 14 | 3,752 |
Tags: dp
Correct Solution:
```
#!/usr/bin/env python3
import os
import sys
from io import BytesIO, IOBase
class FastO:
def __init__(self, fd=1):
stream = BytesIO()
self.flush = lambda: os.write(fd, stream.getvalue()) and not stream.truncate(0) and stream.seek(0)
self.write = lambda b: stream.write(b.encode())
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
sys.stdout, cout = FastO(), ostream()
numbers, num, sign = [], 0, 1
for char in os.read(0, os.fstat(0).st_size):
if char >= 48:
num = num * 10 + char - 48
elif char == 45:
sign = -1
elif char != 13:
numbers.append(sign*num)
num, sign = 0, 1
if char >= 48:
numbers.append(sign*num)
getnum = numbers[::-1].pop
def main():
n, m = getnum(), getnum()
_dp = [[] for _ in range(n)]
for i in range(n):
ni = getnum()
ai = [getnum() for _ in range(ni)]
cumsum = [0] * (ni + 1)
for j in range(ni):
cumsum[j + 1] = cumsum[j] + ai[j]
_dpi = [0] * (ni + 1)
for j in range(ni + 1):
for k in range(j, ni + 1):
_dpi[ni + j - k] = max(_dpi[ni + j - k], cumsum[j] + cumsum[-1] - cumsum[k])
_dp[i] = _dpi
dp = [0] * (m + 1)
for i in range(n):
_dpi = _dp[i]
for j in reversed(range(m + 1)):
dpj = dp[j]
for k in range(min(len(_dp[i]), j + 1)):
dpj = max(dpj, _dpi[k] + dp[j - k])
dp[j] = dpj
cout << dp[m]
if __name__ == '__main__':
main()
``` | output | 1 | 1,876 | 14 | 3,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there n pupils who sat at the <image> desks, two people at each desk. Vasily quickly realized that number n is even. Like all true mathematicians, Vasily has all students numbered from 1 to n.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input
The first input line contains a single even integer n (4 ≤ n ≤ 100) — the number of students in the class. The second line contains exactly n capital English letters "L" and "R". If the i-th letter at the second line equals "L", then the student number i is a lefthander, otherwise he is a righthander.
Output
Print <image> integer pairs, one pair per line. In the i-th line print the numbers of students that will sit at the i-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Examples
Input
6
LLRLLL
Output
1 4
2 5
6 3
Input
4
RRLL
Output
3 1
4 2
Submitted Solution:
```
def main():
with open('input.txt', 'r') as f:
n, s = int(f.readline()), f.readline()
with open('output.txt', 'w') as f:
for i in range(0, n, 2):
j = i + n // 2
if s[i] == 'R' and s[j] == 'L':
i, j = j, i
f.write("%d %d\n" % (i + 1, j + 1))
if __name__ == '__main__':
main()
``` | instruction | 0 | 1,914 | 14 | 3,828 |
No | output | 1 | 1,914 | 14 | 3,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there n pupils who sat at the <image> desks, two people at each desk. Vasily quickly realized that number n is even. Like all true mathematicians, Vasily has all students numbered from 1 to n.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input
The first input line contains a single even integer n (4 ≤ n ≤ 100) — the number of students in the class. The second line contains exactly n capital English letters "L" and "R". If the i-th letter at the second line equals "L", then the student number i is a lefthander, otherwise he is a righthander.
Output
Print <image> integer pairs, one pair per line. In the i-th line print the numbers of students that will sit at the i-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Examples
Input
6
LLRLLL
Output
1 4
2 5
6 3
Input
4
RRLL
Output
3 1
4 2
Submitted Solution:
```
i,o=open("input.txt","r"),open("output.txt","w")
n,a=int(i.readline().strip()),i.readline().strip()
for i in range(n-2) :
if a[i] == 'L' :
o.write(str(i+1)+" "+str(i+3)+"\n")
else :
o.write(str(i+3)+" "+str(i+1)+"\n")
``` | instruction | 0 | 1,915 | 14 | 3,830 |
No | output | 1 | 1,915 | 14 | 3,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there n pupils who sat at the <image> desks, two people at each desk. Vasily quickly realized that number n is even. Like all true mathematicians, Vasily has all students numbered from 1 to n.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input
The first input line contains a single even integer n (4 ≤ n ≤ 100) — the number of students in the class. The second line contains exactly n capital English letters "L" and "R". If the i-th letter at the second line equals "L", then the student number i is a lefthander, otherwise he is a righthander.
Output
Print <image> integer pairs, one pair per line. In the i-th line print the numbers of students that will sit at the i-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Examples
Input
6
LLRLLL
Output
1 4
2 5
6 3
Input
4
RRLL
Output
3 1
4 2
Submitted Solution:
```
#import math
#n, m = input().split()
#n = int (n)
#m = int (m)
#n, m, k , l= input().split()
#n = int (n)
#m = int (m)
#k = int (k)
#l = int(l)
#n = int(input())
#m = int(input())
#s = input()
##t = input()
#a = list(map(char, input().split()))
#a.append('.')
#print(l)
#c = list(map(int, input().split()))
#c = sorted(c)
#x1, y1, x2, y2 =map(int,input().split())
#n = int(input())
#f = []
#t = [0]*n
#f = [(int(s1[0]),s1[1]), (int(s2[0]),s2[1]), (int(s3[0]), s3[1])]
#f1 = sorted(t, key = lambda tup: tup[0])
inp=open("input.txt","r")
out = open("output.txt","w")
n = int(inp.readline())
s = inp.readline()
l = 0
r = 0
sitz = [0] * n
for i in range (n):
if(s[i] == 'R' and r < n/2):
sitz[r*2+1] = i
r += 1
elif (s[i] == 'R'):
sitz[l*2] = i
l += 1
elif(s[i] == 'L' and l < n/2):
sitz[l*2] = i
l += 1
elif (s[i] == 'L'):
sitz[r*2+1] = i
r += 1
for i in range(0,n, 2):
out.write(str(sitz[i]+1) + " "+ str(sitz[i+1]+1)+ "\n")
out.close()
inp.close()
``` | instruction | 0 | 1,916 | 14 | 3,832 |
No | output | 1 | 1,916 | 14 | 3,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fine October day a mathematics teacher Vasily Petrov went to a class and saw there n pupils who sat at the <image> desks, two people at each desk. Vasily quickly realized that number n is even. Like all true mathematicians, Vasily has all students numbered from 1 to n.
But Vasily Petrov did not like the way the children were seated at the desks. According to him, the students whose numbers differ by 1, can not sit together, as they talk to each other all the time, distract others and misbehave.
On the other hand, if a righthanded student sits at the left end of the desk and a lefthanded student sits at the right end of the desk, they hit elbows all the time and distract each other. In other cases, the students who sit at the same desk, do not interfere with each other.
Vasily knows very well which students are lefthanders and which ones are righthanders, and he asks you to come up with any order that meets these two uncomplicated conditions (students do not talk to each other and do not bump their elbows). It is guaranteed that the input is such that at least one way to seat the students always exists.
Input
The first input line contains a single even integer n (4 ≤ n ≤ 100) — the number of students in the class. The second line contains exactly n capital English letters "L" and "R". If the i-th letter at the second line equals "L", then the student number i is a lefthander, otherwise he is a righthander.
Output
Print <image> integer pairs, one pair per line. In the i-th line print the numbers of students that will sit at the i-th desk. The first number in the pair stands for the student who is sitting to the left, and the second number stands for the student who is sitting to the right. Separate the numbers in the pairs by spaces. If there are multiple solutions, print any of them.
Examples
Input
6
LLRLLL
Output
1 4
2 5
6 3
Input
4
RRLL
Output
3 1
4 2
Submitted Solution:
```
#import math
#n, m = input().split()
#n = int (n)
#m = int (m)
#n, m, k , l= input().split()
#n = int (n)
#m = int (m)
#k = int (k)
#l = int(l)
#n = int(input())
#m = int(input())
#s = input()
##t = input()
#a = list(map(char, input().split()))
#a.append('.')
#print(l)
#c = list(map(int, input().split()))
#c = sorted(c)
#x1, y1, x2, y2 =map(int,input().split())
#n = int(input())
#f = []
#t = [0]*n
#f = [(int(s1[0]),s1[1]), (int(s2[0]),s2[1]), (int(s3[0]), s3[1])]
#f1 = sorted(t, key = lambda tup: tup[0])
inp=open("input.txt","r")
out = open("output.txt","w")
n = int(inp.readline())
s = inp.readline()
l = 0
r = 0
sitz = [0] * n
for i in range (n):
if l < n/2:
sitz[l *2] = i
l += 1
else:
sitz[r*2+1] = i
r += 1
for i in range(0,n, 2):
if (s[sitz[i] == 'L' and s[sitz[i+1]] == 'R']):
sitz[i], sitz[i+1] = sitz[i+1], sitz[i]
out.write(str(sitz[i]+1) + " "+ str(sitz[i+1]+1)+ "\n")
out.close()
inp.close()
``` | instruction | 0 | 1,917 | 14 | 3,834 |
No | output | 1 | 1,917 | 14 | 3,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.
The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbers l and r such that r - l + 1 is even. After that animals that occupy positions between l and r inclusively are rearranged as follows: the animal at position l swaps places with the animal at position l + 1, the animal l + 2 swaps with the animal l + 3, ..., finally, the animal at position r - 1 swaps with the animal r.
Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most 20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the height of the animal occupying the i-th place.
Output
Print the sequence of operations that will rearrange the animals by non-decreasing height.
The output should contain several lines, i-th of the lines should contain two space-separated integers li and ri (1 ≤ li < ri ≤ n) — descriptions of segments the robber girl should name. The segments should be described in the order the operations are performed.
The number of operations should not exceed 20 000.
If the animals are arranged correctly from the start, you are allowed to output nothing.
Examples
Input
4
2 1 4 3
Output
1 4
Input
7
36 28 57 39 66 69 68
Output
1 4
6 7
Input
5
1 2 1 2 1
Output
2 5
3 4
1 4
1 4
Note
Note that you don't have to minimize the number of operations. Any solution that performs at most 20 000 operations is allowed. | instruction | 0 | 2,098 | 14 | 4,196 |
Tags: constructive algorithms, implementation, sortings
Correct Solution:
```
def zoo(lst):
result = list()
for i in range(len(lst) + 2):
for j in range(len(lst) - 1):
if lst[j] > lst[j + 1]:
result.append([j + 1, j + 2])
lst[j], lst[j + 1] = lst[j + 1], lst[j]
return result
n = int(input())
a = [int(x) for x in input().split()]
for elem in zoo(a):
print(*elem)
``` | output | 1 | 2,098 | 14 | 4,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,167 | 14 | 4,334 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n = int(input())
l = [int(x) for x in input().split()]
max_claw = 0
alive_count = 1
for i in range(n-1, -1, -1):
max_claw = max(max_claw, l[i])
if max_claw == 0 and i > 0:
alive_count += 1
else:
max_claw -= 1
print(alive_count)
``` | output | 1 | 2,167 | 14 | 4,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,168 | 14 | 4,336 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
kill = 0
minus = 0
s = []
for i in range(n - 1, -1, -1):
if kill > 0:
kill -= 1;
minus += 1;
kill = max(kill, a[i])
print(n - minus)
``` | output | 1 | 2,168 | 14 | 4,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,169 | 14 | 4,338 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
if __name__ == "__main__":
n = int(input())
t = list(map(int, input().split()))
a = [0] * n
for i in range(n):
a[max(0, i - t[i])] += 1
a[i] -= 1
res = 0
for i in range(n):
if i > 0:
a[i] += a[i - 1]
if a[i] == 0:
res += 1
print(res)
``` | output | 1 | 2,169 | 14 | 4,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,170 | 14 | 4,340 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
q=int(input())
a=list(map(int,input().split()))
ans=0
t=q
for i in range(q-1,-1,-1):
if t>i:
ans+=1
t=min(t,i-a[i])
print(ans)
``` | output | 1 | 2,170 | 14 | 4,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,171 | 14 | 4,342 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n = int(input())
claws = list(map(int, input().split()))
c = 0
r = n
for claw in reversed(claws):
if c > 0:
r -= 1
c = max(c - 1, claw)
print(r)
``` | output | 1 | 2,171 | 14 | 4,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,172 | 14 | 4,344 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
#10
#1 1 3 0 0 0 2 1 0 3
n = int(input())
lv = list(map(int,input().strip().split(' ')))
ans = 0
safe = n
i = n
while i > 0:
if i <= safe:
ans += 1
safe = min(safe,i - lv[i-1] - 1)
if safe < 0:
break
i -= 1
print(ans)
``` | output | 1 | 2,172 | 14 | 4,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,173 | 14 | 4,346 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
#Wrath
n = int(input())
l = list(map(int, input().split()))
i = n - 1
j = n - 1
count = 0
while j >= 0 and i >= 0:
if j >= i:
count += 1
j = i - l[i] - 1
if i - l[i] <= j:
j = i - l[i] - 1
i -= 1
print(count)
``` | output | 1 | 2,173 | 14 | 4,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him. | instruction | 0 | 2,174 | 14 | 4,348 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
left = n - 1
right = n - 1
ans = 0
for i in range(n-1, -1, -1):
right = i
if left == right:
ans += 1
left = min(left, right - l[i] - 1)
print(ans)
``` | output | 1 | 2,174 | 14 | 4,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n = int(input())
l = list(map(int,input().split()))
s = set()
i = len(l)-1
j = len(l)-2
while(i > -1):
while(j > -1 and i > j):
if (j+1) >= ((i+1)-l[i]):
#print(i,j)
s.add(j)
j = j-1
else:
break
if i == j:
j = j-1
else:
i = i-1
print(len(l)-len(s))
``` | instruction | 0 | 2,175 | 14 | 4,350 |
Yes | output | 1 | 2,175 | 14 | 4,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
def solve(n,seq) :
alive = n
pointer = n-1
alivePointer = n-2
while pointer > 0 and alivePointer >= 0 :
value = (pointer+1) - seq[pointer]
if value <= 0 :
value = 1
value -= 1
if alivePointer == pointer :
alivePointer -= 1
if alivePointer >= value :
diff = alivePointer - value + 1
alivePointer = value - 1
alive -= diff
pointer -= 1
return alive
n = int(input())
seq = list(map(int,input().split()))
print (solve(n,seq))
``` | instruction | 0 | 2,176 | 14 | 4,352 |
Yes | output | 1 | 2,176 | 14 | 4,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n = int(input())
L = [int(i) for i in input().split()]
pomoika = 0
l = n
for i in range(n - 1, 0, -1):
nl = max(i - L[i], 0)
if l > nl:
if i < l:
pomoika += i - nl
else:
pomoika += l - nl
l = nl
if l <= 0:
break
print(n - pomoika)
``` | instruction | 0 | 2,177 | 14 | 4,354 |
Yes | output | 1 | 2,177 | 14 | 4,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n = int(input())
a = [i - int(x) + 1 for i, x in enumerate(input().split())]
a.insert(0, 0)
min_a = a[-1]
res = n
for i in range(n - 1, 0, -1):
if i >= min_a:
res -= 1
min_a = min(min_a, a[i])
print(res)
``` | instruction | 0 | 2,178 | 14 | 4,356 |
Yes | output | 1 | 2,178 | 14 | 4,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n=int(input())
l=input().split()
count=1
var=0
j=1
while j<n:
if l[n-j]=='0':
count+=1
j+=1
else:
j=j+int(l[n-j])
print(count)
``` | instruction | 0 | 2,179 | 14 | 4,358 |
No | output | 1 | 2,179 | 14 | 4,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n=int(input())
m=list(map(int,input().split()))
c=set()
x=1
for i in range(len(m)):
x-=1
for j in range(1,min(i,m[i]-x)+1):
c.add(i-j)
x=max(x,m[i],1)
print(n-len(c))
``` | instruction | 0 | 2,180 | 14 | 4,360 |
No | output | 1 | 2,180 | 14 | 4,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
size = int(input())
array = [int(i) for i in input().split()]
check = [0]*size
total = 0
for i in range(size-1,0,-1):
if array[i]:
val = i - array[i] if 0 < i - array[i] else 0
check[val] = 1
pos = size -1
for i in range(size-1,-1,-1):
if check[i]:
total += pos - i
for j in range(i,-1,-1):
if array[j]:
pos = j
break
print(size-total)
``` | instruction | 0 | 2,181 | 14 | 4,362 |
No | output | 1 | 2,181 | 14 | 4,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hands that shed innocent blood!
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input
The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.
Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.
Output
Print one integer — the total number of alive people after the bell rings.
Examples
Input
4
0 1 0 10
Output
1
Input
2
0 0
Output
2
Input
10
1 1 3 0 0 0 2 1 0 3
Output
3
Note
In first sample the last person kills everyone in front of him.
Submitted Solution:
```
n=int(input())
claw=[int(x) for x in input().split()]
kill=[0]*n
for i in range(n-2,-1,-1):
kill[i]=claw[i+1]+max(kill[i+1]-1,0)
alive=0
for i in range(n):
if kill[i]<2:
alive+=1
print(alive)
``` | instruction | 0 | 2,182 | 14 | 4,364 |
No | output | 1 | 2,182 | 14 | 4,365 |
Provide a correct Python 3 solution for this coding contest problem.
Joisino is working as a receptionist at a theater.
The theater has 100000 seats, numbered from 1 to 100000.
According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive).
How many people are sitting at the theater now?
Constraints
* 1≤N≤1000
* 1≤l_i≤r_i≤100000
* No seat is occupied by more than one person.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
l_1 r_1
:
l_N r_N
Output
Print the number of people sitting at the theater.
Examples
Input
1
24 30
Output
7
Input
2
6 8
3 3
Output
4 | instruction | 0 | 2,319 | 14 | 4,638 |
"Correct Solution:
```
n = int(input())
cnt = 0
for i in range(n):
l,r = map(int,input().split());cnt += r-l+1
print(cnt)
``` | output | 1 | 2,319 | 14 | 4,639 |
Provide a correct Python 3 solution for this coding contest problem.
Joisino is working as a receptionist at a theater.
The theater has 100000 seats, numbered from 1 to 100000.
According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive).
How many people are sitting at the theater now?
Constraints
* 1≤N≤1000
* 1≤l_i≤r_i≤100000
* No seat is occupied by more than one person.
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N
l_1 r_1
:
l_N r_N
Output
Print the number of people sitting at the theater.
Examples
Input
1
24 30
Output
7
Input
2
6 8
3 3
Output
4 | instruction | 0 | 2,320 | 14 | 4,640 |
"Correct Solution:
```
N=int(input())
ans=0
for i in range(N):
l,r=map(int,input().split())
ans+=abs(l-r)+1
print(ans)
``` | output | 1 | 2,320 | 14 | 4,641 |
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