message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3. | instruction | 0 | 40,730 | 22 | 81,460 |
Tags: implementation, math, number theory
Correct Solution:
```
from math import gcd
p=int(input())-1
cnt=0
for i in range(1,p):
if gcd(i,p)==1: cnt+=1
print(cnt if p!=1 else 1)
``` | output | 1 | 40,730 | 22 | 81,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3. | instruction | 0 | 40,731 | 22 | 81,462 |
Tags: implementation, math, number theory
Correct Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
ORDA = 97 # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=2):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
new_number = 0
while number > 0:
new_number += number % base
number //= base
return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
for i in range(len(s) // 2 + 1):
if s[i] != s[-i - 1]:
return False
return True
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way
#gcd
#for i in range(1,n):
# gcdx = math.gcd(gcdx,arr[i])
p = ii()
#r = sieve(p)
phi = []
for i in range(1,p):
if math.gcd(i,p) == 1:
phi.append(i)
#print(phi)
phiofphi = len(phi)
r = sieve(phiofphi)
for i in r:
if phiofphi%i == 0:
phiofphi = phiofphi*(i-1)//i
print(int(phiofphi))
``` | output | 1 | 40,731 | 22 | 81,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=1):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
newnumber = 0
while number > 0:
newnumber += number % base
number //= base
return newnumber
def cdiv(n, k): return n // k + (n % k != 0)
p = ii()
ans = 0
for i in range(1, p, 1):
t = ''
for j in range(1, p - 1, 1):
if pow(i, j, p) != 1:
t += '0'
else:
break
if pow(i, p - 1, p) == 1:
t += '1'
if t == '0' * (p - 2) + '1':
ans += 1
print(ans)
``` | instruction | 0 | 40,732 | 22 | 81,464 |
Yes | output | 1 | 40,732 | 22 | 81,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
x=int(input())
if(x==2):
print(1)
else:
x-=1
cnt=0
for i in range(1,x):
ok=0
for j in range(2,i+1):
if(x%j==0 and i%j==0): ok=1
if(ok==0) :cnt+=1
print(cnt)
``` | instruction | 0 | 40,733 | 22 | 81,466 |
Yes | output | 1 | 40,733 | 22 | 81,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
t=0
from fractions import gcd
x=int(input())-1
for i in range(1, x+1):
if gcd(x, i)==1:
t+=1
print(t)
``` | instruction | 0 | 40,734 | 22 | 81,468 |
Yes | output | 1 | 40,734 | 22 | 81,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
import math
sol = 0
p=int(input())
for e in range(1, p):
if math.gcd(p-1, e) == 1:
sol+=1
print(sol)
``` | instruction | 0 | 40,735 | 22 | 81,470 |
Yes | output | 1 | 40,735 | 22 | 81,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
mod=10**9+7
# sys.setrecursionlimit(10**6)
def main():
p=int(input())
ans=0
for i in range(2,p):
temp=1
for _ in range(p-2):
temp*=i
temp%=p
if temp==1:
break
else:
ans+=1
print(ans)
#----------------------------------------------------------------------------------------
def nouse0():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse1():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse2():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# region fastio
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 nouse3():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse4():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse5():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# endregion
if __name__ == '__main__':
main()
``` | instruction | 0 | 40,736 | 22 | 81,472 |
No | output | 1 | 40,736 | 22 | 81,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
n=int(input())
count=0
for i in range(1,n):
if((i**(n-1)-1)%n==0):
t=True
for j in range(2,n):
b=i**(n-j)-1
if(b<n):
break
if(b%n==0):
t=False
break
if(t):
count+=1
print(count)
``` | instruction | 0 | 40,737 | 22 | 81,474 |
No | output | 1 | 40,737 | 22 | 81,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
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])
bo = [True]*n
bo [0] = False
bo[1] = False
for i in range(2,n):
if bo[i]:
j = 2
while(i*j < n):
bo[i*j] = False
j += 1
c = 0
for i in range (0, n):
if bo[i]:
x = i
for j in range(1, n-2):
if (x-1) % n == 0:
bo[i] = False
break
x *=x
if (x -1) % n != 0:
bo[i] = False
if bo[i]:
c += 1
if (n != 3):
print(c)
else:
print(1)
``` | instruction | 0 | 40,738 | 22 | 81,476 |
No | output | 1 | 40,738 | 22 | 81,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.
Input
The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.
Output
Output on a single line the number of primitive roots <image>.
Examples
Input
3
Output
1
Input
5
Output
2
Note
The only primitive root <image> is 2.
The primitive roots <image> are 2 and 3.
Submitted Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=1):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
newnumber = 0
while number > 0:
newnumber += number % base
number //= base
return newnumber
def cdiv(n, k): return n // k + (n % k != 0)
p = ii()
ans = 0
for i in range(2, p, 1):
t = ''
for j in range(1, p, 1):
if pow(i, j, p) == 1:
t += '1'
else:
t += '0'
if t == '0' * (p - 2) + '1':
ans += 1
print(ans)
``` | instruction | 0 | 40,739 | 22 | 81,478 |
No | output | 1 | 40,739 | 22 | 81,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,820 | 22 | 81,640 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split())
a, b = 1, 0
for i in range(50):
c, d = s & (1 << i), x & (1 << i)
# print(i, a, b, c, d)
if c == d:
if c:
a, b = 2 * a, 0
else:
a, b = a, a
else:
if c:
a, b = b, b
else:
a, b = 0, 2 * b
if s == x:
a -= 2
print(a)
``` | output | 1 | 40,820 | 22 | 81,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,821 | 22 | 81,642 |
Tags: dp, math
Correct Solution:
```
def solve(s, x):
d = (s - x)
if x << 1 & d or d%2 or d<0: return 0
return 2 ** (bin(x).count('1')) - (0 if d else 2)
s, x = [int(x) for x in input().split()]
print(solve(s, x))
``` | output | 1 | 40,821 | 22 | 81,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,822 | 22 | 81,644 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split())
print(0 if s < x or (s - x) & (2 * x + 1) else 2 ** bin(x).count('1') - 2 * (s == x))
``` | output | 1 | 40,822 | 22 | 81,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,823 | 22 | 81,646 |
Tags: dp, math
Correct Solution:
```
s, x = map(int, input().split())
rem = int(s == x) * 2
p, t, cur = [], 0, 1
for i in range(64):
if x % 2:
t += 1
s -= cur
else:
p.append(cur * 2)
cur *= 2
x //= 2
for i in p[::-1]:
if s >= i: s -= i
ans = 0 if s else 2 ** t - rem
print(ans)
``` | output | 1 | 40,823 | 22 | 81,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,824 | 22 | 81,648 |
Tags: dp, math
Correct Solution:
```
a, b = [int(x) for x in input().split()]
c = (a-b) / 2
if c < 0 or not c.is_integer() or int(c) & b:
print(0)
exit(0)
t = 0
while b:
t += b & 1
b >>= 1
t = 1 << t
if c == 0:
t -= 2
print(t)
``` | output | 1 | 40,824 | 22 | 81,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,825 | 22 | 81,650 |
Tags: dp, math
Correct Solution:
```
a, b = [int(x) for x in input().split()]
c = (a-b) / 2
if not c.is_integer() or int(c) & b:
print(0)
exit(0)
t = 0
while b: # for each x_i, y_i in binary only 0, 1 or 1, 0 valid if b_i == 1
t += b & 1
b >>= 1
t = 1 << t
if c == 0: # for x = 0, y = a or swapped
t -= 2
print(t)
``` | output | 1 | 40,825 | 22 | 81,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,826 | 22 | 81,652 |
Tags: dp, math
Correct Solution:
```
s,x=map(int,input().split())
f=0
if s==x:
f=-2
aa=s-x
if aa%2==1 or aa<0:
print(0)
else:
a=aa//2
#print(a,s)
out=1
for i in range(64):
xx=x%2
aa=a%2
if xx==1:
out*=2
if aa==1:
out=0
x//=2
a//=2
print(out+f)
``` | output | 1 | 40,826 | 22 | 81,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1). | instruction | 0 | 40,827 | 22 | 81,654 |
Tags: dp, math
Correct Solution:
```
import io
import sys
import time
import random
#~ start = time.clock()
#~ test = '''3 3''' # 2 solutions, (1,2) and (2,1)
#~ test = '''9 5''' # 4 solutions
#~ test = '''5 2''' # 0 solutions
#~ test = '''6 0''' # 1 solution
#~ sys.stdin = io.StringIO(test)
s,x = list(map(int, input().split()))
#~ print(s,x)
bitlen = s.bit_length()
def bits_of(x,bitlen):
return [int((1<<i)&x!=0) for i in range(bitlen)]
sbits = bits_of(s,bitlen)
xbits = bits_of(x,bitlen)
overflows = bits_of(s^x,bitlen+1)
#~ print("s",sbits)
#~ print("x",xbits)
#~ print("overflows",overflows)
count = 1
if overflows[0]!=0 or s<x:
count = 0
else:
zero_is_solution = True
for i in range(bitlen):
sumof_a_and_b = 2*overflows[i+1]+sbits[i]-overflows[i]
#~ print(i,":",xbits[i],overflows[i+1],sbits[i],sumof_a_and_b)
if (sumof_a_and_b==0 and xbits[i]==1) or \
(sumof_a_and_b==1 and xbits[i]==0) or \
(sumof_a_and_b==2 and xbits[i]==1) or \
sumof_a_and_b>2 or sumof_a_and_b<0:
count = 0
break
if sumof_a_and_b==1 and xbits[i]==1:
count *= 2
#~ if sumof_a_and_b==0 and xbits[i]==0:
#~ print("s",(0,0))
#~ if sumof_a_and_b==1 and xbits[i]==1:
#~ print("s",(0,1),(1,0))
if sumof_a_and_b==2 and xbits[i]==0:
#~ print("s",(1,1))
zero_is_solution = False
if count>0 and zero_is_solution:
#~ print("R")
count -= 2
print(count)
#~ dur = time.clock()-start
#~ print("Time:",dur)
``` | output | 1 | 40,827 | 22 | 81,655 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,942 | 22 | 81,884 |
Tags: math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
def inverse(a,m):
return pow(a,m-2,m)
def array():
l=raw_input().strip()
arr=[]
for i in l:
if i=='+':
arr.append(1)
else:
arr.append(-1)
return arr
range = xrange # not for python 3.0+
mod=10**9+9
n,a,b,k=in_arr()
l=array()
ans=0
cm=(b*inverse(a,mod))%mod
val=pow(cm,k,mod)
x=(n+1)/k
if val>1:
mul=((pow(val,x,mod)-1)*inverse(val-1,mod))%mod
else:
mul=x
temp=pow(a,n,mod)
for i in range(k):
ans=(ans+(l[i]*temp*mul))%mod
temp=(temp*cm)%mod
pr_num(ans)
``` | output | 1 | 40,942 | 22 | 81,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,943 | 22 | 81,886 |
Tags: math, number theory
Correct Solution:
```
import sys
MOD = 10**9 + 9
def inv(x):
return pow(x, MOD-2, MOD)
def alternateSum(n,a,b,k,s):
res = 0
q = (pow(b, k, MOD) * inv(pow(a, k, MOD))) % MOD
max_pow = pow(a, n, MOD)
c = b * inv(a) % MOD
for i in range(k):
if s[i] == '+':
res += max_pow
elif s[i] == '-':
res -= max_pow
res %= MOD
max_pow = (max_pow * c) % MOD
t = (n+1) // k
if q == 1:
return (t*res) % MOD
z = ((pow(q, t, MOD) - 1) * inv(q-1)) % MOD
return z * res % MOD
n, a, b, k, s = sys.stdin.read().split()
result = alternateSum(int(n), int(a), int(b), int(k), s)
print(result)
``` | output | 1 | 40,943 | 22 | 81,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,944 | 22 | 81,888 |
Tags: math, number theory
Correct Solution:
```
MOD = 1000000009
def Pow(base, n):
res = 1
while n:
if n&1:
res = (res*base)%MOD
base = (base*base)%MOD
n >>= 1
return res
n, a, b, k = map(int, input().split())
ans = 0
num = (n+1)//k
_a = Pow(a, MOD-2)
q = Pow(b, k)*Pow(Pow(a, k), MOD-2)%MOD
if q == 1:
res = Pow(a, n)*num%MOD
for i in input():
if i == '+':
ans = (ans+res)%MOD
else:
ans = (ans-res)%MOD
res = res*b%MOD*_a%MOD
else:
# rat = (1-Pow(q, num))%MOD*Pow((1-q)%MOD, MOD-2)%MOD
rat = (Pow(q, num)-1)%MOD*Pow((q-1)%MOD, MOD-2)%MOD
cur = Pow(a, n)*rat%MOD
for i in input():
if i == '+':
ans = (ans+cur)%MOD
else:
ans = (ans-cur)%MOD
cur = cur*b%MOD*_a%MOD
print(ans)
``` | output | 1 | 40,944 | 22 | 81,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,945 | 22 | 81,890 |
Tags: math, number theory
Correct Solution:
```
import functools
import queue
def readTuple():
return input().split()
def readInts():
return tuple(map(int, readTuple()))
def egcd(aa, bb):
lastremainder, remainder = abs(aa), abs(bb)
x, lastx, y, lasty = 0, 1, 1, 0
while remainder:
lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
x, lastx = lastx - quotient*x, x
y, lasty = lasty - quotient*y, y
return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise AtributeError(a,m)
else:
return x % m
@functools.lru_cache(None)
def mpow(b, exp, m):
if exp == 0:
return 1
if exp%2:
return (b*mpow(b,exp-1,m))%m
return (mpow(b, exp//2, m)**2)%m
# r = 1
# while exp>0:
# if exp%2:
# r = (r*b)%m
# b = (b*b)%m
# exp = exp//2
# return r%m
def solve():
MOD = 10**9 +9
n, a, b, k = readInts()
vals = input().strip()
base = 0
BB = 1
for i,v in enumerate(vals):
if v == '+':
base += mpow(a,n-i,MOD)*mpow(b,i, MOD)
else:
base -= mpow(a,n-i,MOD)*mpow(b,i, MOD)
base %= MOD
l = (n+1)//k
xx = mpow(modinv(a,MOD)*b, k, MOD)
if xx != 1:
mul = (1-mpow(xx, l, MOD))*modinv(1-xx, MOD) %MOD
else:
mul = l
print((base*mul)%MOD)
if __name__ == '__main__':
solve()
``` | output | 1 | 40,945 | 22 | 81,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,946 | 22 | 81,892 |
Tags: math, number theory
Correct Solution:
```
import sys
try:
fin = open('in')
except:
fin = sys.stdin
input = fin.readline
mod=10**9+9
def f(x,y):
ans=1
while y:
if y&1:ans=ans*x%mod
x=x*x%mod
y>>=1
return ans
n,a,b,k=map(int,input().split())
s=[1 if c=='+' else -1 for c in input()]
#period k-1
pr=sum(s[i]*f(a,n-i)*f(b,i) for i in range(k))%mod
#ratio (b/a)^k
rt=f(b,k)*f(f(a,k),mod-2)%mod
terms=(n+1)//k
if rt==1:
print(terms*pr%mod)
else:
print(pr*(f(rt,terms)-1)*f(rt-1,mod-2)%mod)
``` | output | 1 | 40,946 | 22 | 81,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,947 | 22 | 81,894 |
Tags: math, number theory
Correct Solution:
```
MOD = 1000000009
def inv(n):
return pow(n, MOD - 2, MOD)
n, a, b, k = map(int, input().split())
q = (n + 1) // k
string = input()
s = []
for char in string:
if char == "+": s.append(1)
else: s.append(-1)
res = 0 # final answer
for i in range(k):
res += (s[i] * pow(a, n - i, MOD) * pow(b, i, MOD))
res %= MOD
n1 = pow(b, k, MOD)
n2 = pow(a, k, MOD)
n2 = inv(n2)
T = n1 * n2 % MOD
if a != b and T != 1:
num = (pow(b, n + 1, MOD) - pow(a, n + 1, MOD)) % MOD # numerator
d1 = (pow(b, k, MOD) - pow(a, k, MOD)) % MOD
d1 = inv(d1)
d2 = pow(a, n + 1 - k, MOD)
d2 = inv(d2)
R = (num * d1 * d2) % MOD
res = (res * R) % MOD
print(res)
else:
res *= q
res %= MOD
print(res)
``` | output | 1 | 40,947 | 22 | 81,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,948 | 22 | 81,896 |
Tags: math, number theory
Correct Solution:
```
MOD = 10**9+9;
n,a,b,k = map(int,input().split())
s = input();
def pow(a,n,m=MOD):
ret = 1;
a %= MOD;
while n:
if n&1:
ret = ret*a%m;
a = a*a%m;
n >>= 1;
return ret;
#def inv(a,p=MOD):
# return pow(a,p-2,p);
def inv(a,m=MOD):
if a > 1:
return (m-m//a*inv(m%a))%m;
return 1
ia,d = inv(a), (n+1)//k
ans, ci0, q = 0, pow(a,n), pow(ia*b,k)
Q = d
if q != 1:
Q = (pow(q,d)-1)*inv(q-1)%MOD
for i in range(k):
sign = 1
if s[i] == '-':
sign = -1;
ans += sign*ci0*Q%MOD
ans %= MOD
ci0 = ci0*ia*b%MOD
print(ans);
``` | output | 1 | 40,948 | 22 | 81,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,949 | 22 | 81,898 |
Tags: math, number theory
Correct Solution:
```
n , a , b , k = map(int,input().split())
s = str(input())
MOD = 10**9 + 9
def fun(a , b , i):
return (pow(a , n - i , MOD) * pow(b , i , MOD)) % MOD
def div(x , y):
#calculate x // y mod p
return (x * pow(y , MOD - 2 , MOD)) % MOD
r = pow(div(b , a) , k , MOD)
SUM = 0
for i in range(k):
if s[i] == '-':
SUM += -1 * fun(a , b , i)
else:
SUM += +1 * fun(a , b , i)
n += 1
t = n // k
ans = pow(r , t , MOD) - 1
x = r - 1
if x :
ans = (div(ans , x) * SUM) % MOD
else:
ans = (t * SUM) % MOD
ans += MOD
ans %= MOD
print(ans)
``` | output | 1 | 40,949 | 22 | 81,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}. | instruction | 0 | 40,950 | 22 | 81,900 |
Tags: math, number theory
Correct Solution:
```
MOD = int(1e9+9)
n, a, b, k = map(int, input().split())
s = input()
def solve():
res = 0
q = pow(b, k, MOD) * pow(pow(a, k, MOD), MOD-2, MOD) % MOD
max_pow = pow(a, n, MOD)
c = b * pow(a, MOD-2, MOD) % MOD
for i in range(k):
res += max_pow if s[i] == '+' else -max_pow
res = (res % MOD + MOD) % MOD
max_pow = max_pow * c % MOD
t = (n + 1) // k
if q == 1:
return t * res % MOD
z = (pow(q, t, MOD) - 1) * pow(q-1, MOD-2, MOD) % MOD
return z * res % MOD
print(solve())
``` | output | 1 | 40,950 | 22 | 81,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
MOD = 1000000009
def xpow(desk,step):
ret=1
desk=desk%MOD
while step:
if(step%2):
ret=(ret*desk)%MOD
step=step//2
desk = (desk * desk) % MOD
return ret
if __name__ == '__main__':
n,a,b,k=map(int,input().split())
s=input()
base=0
for i in range(0,k):
base = (base + [-1, 1][s[i] == '+'] *(xpow(a, n - i) * xpow(b, i))) % MOD
loop=(n+1)//k
a=xpow(a,k)
b=xpow(b,k)
if(a!=b):
print((((xpow(b,loop)-xpow(a,loop))*xpow((b*xpow(a,loop-1)-xpow(a,loop)),MOD-2))%MOD*base)%MOD)
else:
print((base*loop)%MOD)
``` | instruction | 0 | 40,951 | 22 | 81,902 |
Yes | output | 1 | 40,951 | 22 | 81,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
n, a, b, k = map(int, input().split())
s = input()
m = int(1e9 + 9)
a_1 = pow(a, m - 2, m)
x = (a_1 * b) % m
xk = pow(x, k, m)
# print("xk", xk)
C = 0
for i in range(0, k):
z = 1 if s[i] == "+" else -1
C = (C + z * pow(x, i, m)) % m
# print("C", C)
kk = (n + 1) // k
if xk > 1:
v1 = (pow(xk, kk, m) - 1) % m
v2 = pow( (xk - 1) % m, m - 2, m)
D = (v1 * v2) % m
else:
D = kk
# print("D", D)
ans = pow(a, n, m) * C * D
ans %= m
print(ans)
``` | instruction | 0 | 40,952 | 22 | 81,904 |
Yes | output | 1 | 40,952 | 22 | 81,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
MOD = 1000000009
def xpow(desk,step):
ret=1
desk=desk%MOD
while step:
if(step%2):
ret=(ret*desk)%MOD
step=step//2
desk = (desk * desk) % MOD
return ret
if __name__ == '__main__':
n,a,b,k=map(int,input().split())
s=input()
base=0
for i in range(0,k):
base = (base + [-1, 1][s[i] == '+'] *(xpow(a, n - i) * xpow(b, i))) % MOD
loop=(n+1)//k
a=xpow(a,k)
b=xpow(b,k)
if(a!=b):
print((((xpow(b,loop)-xpow(a,loop))*xpow((b*xpow(a,loop-1)-xpow(a,loop)),MOD-2))*base)%MOD)
else:
print((base*loop)%MOD)
``` | instruction | 0 | 40,953 | 22 | 81,906 |
Yes | output | 1 | 40,953 | 22 | 81,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
raw = input().split()
vals = [x for x in input()]
n,a,b,k = [int(x) for x in raw]
summ = 0
mod = 1000000009
def inv(x):
return fast_power(x, mod-2)
def fast_power(a,n):
ret = 1
a = a % mod
while n:
if n&1:
ret = ret*a%mod
a = a*a%mod
n >>= 1
return ret
c = inv(a) * b % mod
cf = fast_power(c, k)
m = (n + 1) // k
if cf -1:
p = (fast_power(cf, m) - 1) * inv(cf - 1) % mod
else:
p = m
x = fast_power(a, n)
for i in range(k):
summ = (summ + [-1, 1][vals[i] == '+'] * x * p) % mod
x = (x * c) % mod
print(summ)
``` | instruction | 0 | 40,954 | 22 | 81,908 |
Yes | output | 1 | 40,954 | 22 | 81,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
n, a, b, k = input().split()
n, a, b, k = int(n), int(a) % (10 ** 9 + 9), int(b) % (10 ** 9 + 9), int(k)
syms = []
s = input()
for i in s:
if i is '+':
syms.append(1)
else:
syms.append(-1)
temp = 0
ta, tb = (a ** n) % (10 ** 9 + 9), b ** 0
tp = (a ** -1)
for i in range(0, k):
temp += syms[i] * ta * tb
temp %= (10 ** 9 + 9)
ta *= tp
ta %= (10 ** 9 + 9)
tb *= b
tb %= (10 ** 9 + 9)
rst = temp
ta = a ** (-1 * k) % (10 ** 9 + 9)
tb = (b ** k) % (10 ** 9 + 9)
for i in range(1, (n + 1) // k):
temp *= ta * tb
temp %= (10 ** 9 + 9)
rst += temp
rst %= (10 ** 9 + 9)
print(int(rst) % (10 ** 9 + 9))
``` | instruction | 0 | 40,955 | 22 | 81,910 |
No | output | 1 | 40,955 | 22 | 81,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
import atexit
import io
import sys
# Buffering IO
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
def main():
mm = 1000000009
n, a, b, k = [int(x) for x in input().split()]
s = input()
if a ==b:
tt = 0
for c in s:
tt += (1 if c == '+' else -1)
print(tt * pow(a,n,mm) * ((n+1)//k) % mm)
return
"""
if (a < b):
a,b = b,a
s=list(reversed(s))
"""
ss = 0
for i in range(k):
ss = (ss+(pow(a, n+k-i, mm) * pow(b,i,mm) * (1 if s[i] == '+' else -1)))%mm
for i in range(k):
ss = (ss-(pow(b, n+k-i, mm) * pow(a,i,mm) * (1 if s[-i-1] == '+' else -1)))%mm
ttt = (pow(a, k) - pow(b, k))%mm
print ((ss*pow(ttt, mm-2, mm))%mm)
if __name__ == '__main__':
main()
``` | instruction | 0 | 40,956 | 22 | 81,912 |
No | output | 1 | 40,956 | 22 | 81,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
first = str(input())
second = str(input())
List = first.split(' ')
n = int(List[0])
a = int(List[1])
b = int(List[2])
k = int(List[3])
S = second.split(' ')
if len(S) < n+1:
p = ((n+1) / len(S)) + 1
S = S * int(p)
new_S = []
for s in S:
if s == '+':
new_S.append(int(1))
else:
new_S.append(int(-1))
Sum = 0
for i in range(n + 1):
Sum += new_S[i]*(a**(n-i))*(b**i)
print(Sum % 1000000009)
``` | instruction | 0 | 40,957 | 22 | 81,914 |
No | output | 1 | 40,957 | 22 | 81,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.
Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.
Note that the modulo is unusual!
Input
The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).
The second line contains a sequence of length k consisting of characters '+' and '-'.
If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.
Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.
Output
Output a single integer — value of given expression modulo 10^{9} + 9.
Examples
Input
2 2 3 3
+-+
Output
7
Input
4 1 5 1
-
Output
999999228
Note
In the first example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7
In the second example:
(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.
Submitted Solution:
```
import sys
import math
from collections import defaultdict
n,a,b,k=map(int,sys.stdin.readline().split())
s=sys.stdin.readline()[:-1]
#print(s,'s')
#print(n,'n',a,'a',b,'b',k,'k')
mod=10**9+9
ans=0
for i in range(k):
if s[i]=='+':
ans+=pow(a,n-i,mod)*pow(b,i,mod)
ans%=mod
else:
ans-=pow(a,n-i,mod)*pow(b,i,mod)
#print(ans,'before mod')
ans%=mod
#print(ans,'anssss')
x=pow(a,(n+1)//k,mod)*pow(b,(n+1)//k,mod)
if a==b==1:
y=(n+1)//k
else:
y=(x-1)//(a*b-1)
y%=mod
#c=(n+1)//k
#print(c,'c',ans,'ans')
res=((ans)*y)%mod
print(res)
``` | instruction | 0 | 40,958 | 22 | 81,916 |
No | output | 1 | 40,958 | 22 | 81,917 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer n, let us define f(n) as the number of digits in base 10.
You are given an integer S. Count the number of the pairs of positive integers (l, r) (l \leq r) such that f(l) + f(l + 1) + ... + f(r) = S, and find the count modulo 10^9 + 7.
Constraints
* 1 \leq S \leq 10^8
Input
Input is given from Standard Input in the following format:
S
Output
Print the answer.
Examples
Input
1
Output
9
Input
2
Output
98
Input
123
Output
460191684
Input
36018
Output
966522825
Input
1000
Output
184984484 | instruction | 0 | 41,064 | 22 | 82,128 |
"Correct Solution:
```
#!/usr/bin/env python3
import math
import random
class Prime:
seed_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def is_prime(self, n):
is_prime_common = self.is_prime_common(n)
if is_prime_common is not None:
return is_prime_common
if n < 2000000:
return self.is_prime_bf(n)
else:
return self.is_prime_mr(n)
def is_prime_common(self, n):
if n == 1: return False
if n in Prime.seed_primes: return True
if any(map(lambda x: n % x == 0, self.seed_primes)): return False
def is_prime_bf(self, n):
for k in range(2, int(math.sqrt(n)) + 1):
if n % k == 0:
return False
return True
def is_prime_mr(self, n):
d = n - 1
while d % 2 == 0:
d //= 2
witnesses = self.get_witnesses(n)
#witnesses = [random.randint(1, n - 1) for _ in range(100)]
for w in witnesses:
t = d
y = pow(w, t, n)
while t != n - 1 and y != 1 and y != n - 1:
y = (y ** 2) % n
t *= 2
if y != n - 1 and t % 2 == 0:
return False
return True
def get_witnesses(self, num):
def _get_range(num):
if num < 2047:
return 1
if num < 1373653:
return 2
if num < 25326001:
return 3
if num < 3215031751:
return 4
if num < 2152302898747:
return 5
if num < 3474749660383:
return 6
if num < 341550071728321:
return 7
if num < 38255123056546413051:
return 9
return 12
return self.seed_primes[:_get_range(num)]
def gcd(self, a, b):
if a < b:
a, b = b, a
if b == 0:
return a
while b:
a, b = b, a % b
return a
@staticmethod
def f(x, n, seed):
p = Prime.seed_primes[seed % len(Prime.seed_primes)]
return (p * x + seed) % n
def find_factor(self, n, seed=1):
if self.is_prime(n):
return n
x, y, d = 2, 2, 1
count = 0
while d == 1:
count += 1
x = self.f(x, n, seed)
y = self.f(self.f(y, n, seed), n, seed)
d = self.gcd(abs(x - y), n)
if d == n:
return self.find_factor(n, seed+1)
return self.find_factor(d)
def find_factors(self, n):
primes = {}
if self.is_prime(n):
primes[n] = 1
return primes
while n > 1:
factor = self.find_factor(n)
primes.setdefault(factor, 0)
primes[factor] += 1
n //= factor
return primes
def gcd(a, b):
if a < b:
a, b = b, a
while 0 < b:
a, b = b, a % b
return a
def powmod(a, x, m):
y = 1
while 0 < x:
if x % 2 == 1:
y *= a
y %= m
x //= 2
a = a ** 2
a %= M
return y
M = 10 ** 9 + 7
prime = Prime()
def solve(s):
if s == 1:
return 9
ans = 0
n = 1
c = 9
while n * c < s:
n += 1
c *= 10
ans += s // n
for log_r in range(n - 1, n + 1):
c_r = 9 * 10 ** (log_r - 1)
sum_r = log_r * c_r
for log_l in range(1, log_r):
mid_f = 0
for i in range(log_l + 1, log_r):
mid_f += i * 9 * 10 ** (i - 1)
if s <= mid_f:
continue
res = s - mid_f
c_l = 9 * 10 ** (log_l - 1)
if log_l * c_l + sum_r < res:
continue
g = gcd(log_r, log_l)
if res % g != 0:
continue
c_l_max = min(c_l, (res - 1) // log_l)
while 0 < c_l_max:
if (res - log_l * c_l_max) % log_r == 0: break
c_l_max -= 1
if c_l_max == 0:
continue
c_l_min = 1
if sum_r < res:
c_l_min = (res - sum_r + log_l - 1) // log_l
div = log_r // g
ans += (c_l_max - c_l_min + div) // div
factors = prime.find_factors(s)
num_prime_factors = len(factors)
prime_factors = []
count_limit = []
count = [0] * num_prime_factors
for k in factors.keys():
prime_factors.append(k)
count_limit.append(factors[k])
loop = True
while loop:
p = 1
for i, f in enumerate(prime_factors):
p *= f ** count[i]
if n <= p:
ans += 9 * powmod(10, p - 1, M)
ans += M - s // p
ans %= M
count[0] += 1
for i, limit in enumerate(count_limit):
if limit < count[i]:
if i == num_prime_factors - 1:
loop = False
break
count[i + 1] += 1
count[i] = 0
return ans
def main():
s = int(input())
print(solve(s))
if __name__ == '__main__':
main()
``` | output | 1 | 41,064 | 22 | 82,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a positive integer n, let us define f(n) as the number of digits in base 10.
You are given an integer S. Count the number of the pairs of positive integers (l, r) (l \leq r) such that f(l) + f(l + 1) + ... + f(r) = S, and find the count modulo 10^9 + 7.
Constraints
* 1 \leq S \leq 10^8
Input
Input is given from Standard Input in the following format:
S
Output
Print the answer.
Examples
Input
1
Output
9
Input
2
Output
98
Input
123
Output
460191684
Input
36018
Output
966522825
Input
1000
Output
184984484
Submitted Solution:
```
#!/usr/bin/env python3
import math
class Prime:
seed_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def is_prime(self, n):
is_prime_common = self.is_prime_common(n)
if is_prime_common is not None:
return is_prime_common
if n < 2000000:
return self.is_prime_bf(n)
else:
return self.is_prime_mr(n)
def is_prime_common(self, n):
if n == 1: return False
if n in Prime.seed_primes: return True
if any(map(lambda x: n % x == 0, self.seed_primes)): return False
def is_prime_bf(self, n):
for k in range(2, int(math.sqrt(n)) + 1):
if n % k == 0:
return False
return True
def is_prime_mr(self, n):
d = n - 1
while d % 2 == 0:
d //= 2
witnesses = self.get_witnesses(n)
for w in witnesses:
y = pow(w, d, n)
while d != n - 1 and y != 1 and y != n - 1:
y = (y ** 2) % n
d *= 2
if y != n - 1 and d % 2 == 0:
return False
return True
def get_witnesses(self, num):
def _get_range(num):
if num < 2047:
return 1
if num < 1373653:
return 2
if num < 25326001:
return 3
if num < 3215031751:
return 4
if num < 2152302898747:
return 5
if num < 3474749660383:
return 6
if num < 341550071728321:
return 7
if num < 38255123056546413051:
return 9
return 12
return self.seed_primes[:_get_range(num)]
def gcd(self, a, b):
if a < b:
a, b = b, a
if b == 0:
return a
while b:
a, b = b, a % b
return a
@staticmethod
def f(x, n, seed):
p = Prime.seed_primes[seed % len(Prime.seed_primes)]
return (p * x + seed) % n
def find_factor(self, n, seed=1):
if self.is_prime(n):
return n
x, y, d = 2, 2, 1
count = 0
while d == 1:
count += 1
x = self.f(x, n, seed)
y = self.f(self.f(y, n, seed), n, seed)
d = self.gcd(abs(x - y), n)
if d == n:
return self.find_factor(n, seed+1)
return self.find_factor(d)
def find_factors(self, n):
primes = {}
if self.is_prime(n):
primes[n] = 1
return primes
while n > 1:
factor = self.find_factor(n)
primes.setdefault(factor, 0)
primes[factor] += 1
n //= factor
return primes
def gcd(a, b):
if a < b:
a, b = b, a
while 0 < b:
a, b = b, a % b
return a
def powmod(a, x, m):
y = 1
while 0 < x:
if x % 2 == 1:
y *= a
y %= m
x //= 2
a = a ** 2
a %= M
return y
M = 10 ** 9 + 7
prime = Prime()
def solve(s):
if s == 1:
return 9
ans = 0
n = 1
while True:
c = 9 * 10 ** (n - 1)
f = n * c
if s <= f:
break
n += 1
ans += s // n
for log_r in range(n - 1, n + 1):
c_r = 9 * 10 ** (log_r - 1)
sum_r = log_r * c_r
for log_l in range(1, log_r):
mid_f = 0
for i in range(log_l + 1, log_r):
mid_f += i * 9 * 10 ** (i - 1)
if s <= mid_f:
continue
res = s - mid_f
c_l = 9 * 10 ** (log_l - 1)
if log_l * c_l + sum_r < res:
continue
g = gcd(log_r, log_l)
if res % g != 0:
continue
c_l_max = min(c_l, (res - 1) // log_l)
while 0 < c_l_max:
if (res - log_l * c_l_max) % log_r == 0: break
c_l_max -= 1
if c_l_max == 0:
continue
c_l_min = 1
if sum_r < res:
c_l_min = (res - sum_r + log_l - 1) // log_l
div = log_r // g
ans += (c_l_max - c_l_min + div) // div
factors = prime.find_factors(s)
num_prime_factors = len(factors)
prime_factors = []
count_limit = []
count = [0] * num_prime_factors
for k in factors.keys():
prime_factors.append(k)
count_limit.append(factors[k])
loop = True
while loop:
p = 1
for i, f in enumerate(prime_factors):
p *= f ** count[i]
if n <= p:
ans += 9 * powmod(10, p - 1, M)
ans += M - s // p
ans %= M
count[0] += 1
for i, limit in enumerate(count_limit):
if limit < count[i]:
if i == num_prime_factors - 1:
loop = False
break
count[i + 1] += 1
count[i] = 0
return ans
def main():
s = int(input())
print(solve(s))
if __name__ == '__main__':
main()
``` | instruction | 0 | 41,068 | 22 | 82,136 |
No | output | 1 | 41,068 | 22 | 82,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,333 | 22 | 82,666 |
Tags: math, number theory
Correct Solution:
```
x, n= map(int, input().split())
prifa=[]
tag=1
while x>2 and tag:
tag=0
i=2
while i*i<= x:
if x%i==0:
tag=1
while x%i==0:
x//=i
prifa.append(i)
i+=1
if tag:
break;
if x>1:
prifa.append(x)
ans=1
mod=10**9+7
for i in prifa:
mi=0
tmp=n
while tmp:
mi+=tmp//i
tmp//=i
ans*=pow(i,mi,mod)
ans%=mod
print(ans)
``` | output | 1 | 41,333 | 22 | 82,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,334 | 22 | 82,668 |
Tags: math, number theory
Correct Solution:
```
"""
Satwik_Tiwari ;) .
25th AUGUST , 2020 - TUESDAY
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned
#==============================================================================================
#fast I/O region
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
#===============================================================================================
# code here ;))
def solve(case):
x,n = sep()
primes= []
for i in range(1,floor(x**(0.5))+1):
if(x%i==0):
if(isPrime(i)):
primes.append(i)
if((x//i)!=i and isPrime(x//i)):
primes.append(x//i)
# print(primes)
ans = 1
for i in range(len(primes)):
temp = 0
lol = 0
while(primes[i]**(temp+1) <= n):
lol+=n//(primes[i]**(temp+1))
temp+=1
ans*=power(primes[i],lol,mod)
ans%=mod
print(ans%mod)
# for i in range(len(primes)):
# temp = 0
# while(n%(primes[i]**(temp+1)) == 0):
# print(n%(primes[i]**temp),n,primes[i]**temp)
# temp+=1
# primes[i] = [primes[i],temp]
# print(primes)
#
# temp = 1
# for i in range(len(primes)):
# temp*=primes[i][1]+1
# print(temp)
# pow = []
# for i in range(len(primes)):
# lol = (primes[i][1]*(primes[i][1]+1))//2
# print(lol,primes[i],'==')
# pow.append(temp//(primes[i][1]+1) * lol)
# print(pow)
#
# ans = 1
# for i in range(len(pow)):
# ans*=power(primes[i][0],pow[i],mod)
# ans%=mod
#
# print(ans)
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 41,334 | 22 | 82,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,335 | 22 | 82,670 |
Tags: math, number theory
Correct Solution:
```
x, n = map(int, input().split())
m = 10**9 + 7
s = []
for i in range(2, int(x ** .5) + 1):
if x % i == 0:
s.append(i)
while not x % i:
x //= i
if x > 1:
s.append(x)
ans = 1
for i in s:
a = n
res = 0
while (a):
a //= i
res += a
ans *= pow(i, res, m)
print(ans % m)
``` | output | 1 | 41,335 | 22 | 82,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,336 | 22 | 82,672 |
Tags: math, number theory
Correct Solution:
```
x,n = map(int,input().split())
prime_factors = []
for i in range(2, int(x ** 0.5) + 1):
if x % i == 0:
prime_factors.append(i)
while x % i == 0:
x = x // i
if x > 1:
prime_factors.append(x)
ans = 1
mod = int(1e9) + 7
# print(mod)
# print(prime_factors)
for i in prime_factors:
base = i
while base <= n:
ans = (ans * pow(i, n // base, mod)) % mod
base = base * i
# print(ans)
print(ans)
``` | output | 1 | 41,336 | 22 | 82,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,337 | 22 | 82,674 |
Tags: math, number theory
Correct Solution:
```
x, n = [int(i) for i in input().split()]
mod = 1000000007
primes = [] # seznam prafaktorjev
cnt = 2
while cnt * cnt <= x:
if x % cnt == 0:
primes.append(cnt)
while x % cnt == 0:
x = x // cnt
cnt = cnt + 1
if x > 1:
primes.append(x)
ans = 1
for pr in primes:
temp = 0
pk = pr
while pk <= n:
temp = temp + n // pk
pk = pr * pk
ans = ans * pow(pr, temp, mod) % mod
print(ans)
``` | output | 1 | 41,337 | 22 | 82,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,338 | 22 | 82,676 |
Tags: math, number theory
Correct Solution:
```
x,n=map(int,input().split())
ans=1
pp=10**9+7
prime=[]
ss=set()
for i in range(2,int(x**0.5)+1):
if x%i==0:
ss.add(i)
ss.add(x//i)
for d in ss:
i=2
t=True
while t and i<int(d**0.5)+1:
if d%i==0:
t=False
i+=1
if t:
prime.append(d)
if not prime:prime.append(x)
for p in prime:
k=1
r=0
while pow(p,k)<=n:
r+=n//(p**k)
k+=1
ans*=pow(p,r,pp)
print(ans%pp)
``` | output | 1 | 41,338 | 22 | 82,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,339 | 22 | 82,678 |
Tags: math, number theory
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 29 20:05:37 2019
@author: kushagra
"""
import math
p = []
MOD = 1000000007
def gen(n):
if n%2 ==0:
p.append(2)
while n%2==0:
n=n//2
for i in range (3,int(math.sqrt(n))+1,2):
if n%i==0:
p.append(i)
while n%i == 0:
n= n//i
if n>2:
p.append(n)
ll = input().split(" ")
x = int(ll[0])
n = int(ll[1])
gen(x)
ans=1
for i in p:
prime = i
pp = prime
v = []
while 1:
if pp>n:
break
v.append(n//pp)
pp*=prime
v.append(0)
#print(v)
for j in range(0,len(v)-1):
val = int(v[j]-v[j+1])
#val%=MOD-1
x = pow(prime,j+1,MOD)
ans*=pow(x,val,MOD)
print(ans%MOD)
``` | output | 1 | 41,339 | 22 | 82,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue. | instruction | 0 | 41,340 | 22 | 82,680 |
Tags: math, number theory
Correct Solution:
```
from math import sqrt
def primes(n):
L = []
if not (n & 1):
n >>= 1
while not (n & 1):
n >>= 1
L.append(2)
for i in range(3, int(sqrt(n)) + 1, 2):
if n % i == 0:
L.append(i)
n //= i
while n % i == 0:
n //= i
if n > 1:
L.append(n)
return L
def fact_pow(n, p):
ans = 0
while n:
ans += n // p
n //= p
return ans
def binpow(a, n, p):
if n == 0:
return 1
if n == 1:
return a % p
x = binpow(a, n >> 1, p)
return x * x * a % p if n & 1 else x * x % p
ans = 1
mod = 1000000007
x, n = map(int, input().split())
L = primes(x)
for p in L:
ans = (ans * binpow(p, fact_pow(n, p), mod)) % mod
print(ans)
``` | output | 1 | 41,340 | 22 | 82,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue.
Submitted Solution:
```
x,n=input().split(' ');
x=int(x);n=int(n);
mod=int(1e9)+7;
def dec(x):
y=x;i=2; a=[];
while i*i<=x:
if y%i==0:
a.append(i);
while y%i==0: y//=i;
i+=1;
if y>1: a.append(y);
return a
def qpow(a,p):
r=1;
while p:
if p&1: r=r*a%mod;
a=a*a%mod; p>>=1;
return r;
pf=dec(x)
ans=1
for p in pf:
j,qaq,qwq=0,1,p;tmp=0;
while qaq<=n:
t=j*( (n//qaq-n//qwq)%(mod-1) )%(mod-1);
tmp=(tmp+t)%(mod-1);
qaq=qaq*p; qwq=qwq*p; j+=1;
ans=ans*qpow(p,tmp)%mod;
print(ans)
``` | instruction | 0 | 41,341 | 22 | 82,682 |
Yes | output | 1 | 41,341 | 22 | 82,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue.
Submitted Solution:
```
# Prime factorization
def prime_factorization(x):
answer = []
i = 2
while i*i <= x:
if x%i == 0:
answer.append(i)
while x%i == 0: x //= i
i += 1
if x > 1: answer.append(x)
return answer
# Main
def main(X: int, N: int):
answer = 1
mod = 10**9 + 7
x_primes = prime_factorization(x)
for prime in x_primes:
power = 0
factor = prime
while factor <= N:
power += N // factor
factor *= prime
#print("power is:", power)
answer *= pow(prime, power, mod)
answer %= mod
return answer
x, n = [int(c) for c in input().split()]
print(main(x, n))
``` | instruction | 0 | 41,342 | 22 | 82,684 |
Yes | output | 1 | 41,342 | 22 | 82,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue.
Submitted Solution:
```
def power(x, y, p):
res = 1
x = x%p
while (y>0):
if ((y&1) == 1):
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
x,n = map(int,input().split())
r = int(x**0.5)
ans,p = 1,2
primes = set()
while(p<=r and p<=x):
while(x%p==0):
primes.add(p)
x= int(x/p)
p+=1
if(x>1): primes.add(x)
for prime in primes:
e = []
p = prime
while(p<=n):
a = n//p
#print("A ",a)
if(len(e)==0):
e.append(a)
else:
e[len(e)-1] -= a
e.append(a)
p *= prime
total_expo = 0
i = 1
for expo in e:
total_expo += (expo*i)
i+=1
#print(prime,total_expo)
ans = ans*power(prime,total_expo,1000000007)
ans %= 1000000007
print(ans)
``` | instruction | 0 | 41,343 | 22 | 82,686 |
Yes | output | 1 | 41,343 | 22 | 82,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue.
Submitted Solution:
```
def prime_factorization(x):
answer=[]
i=2
while(i*i<=x):
if(x%i==0):
answer.append(i)
while(x%i==0):
x//=i
i+=1
if(x>1):
answer.append(x)
return answer
def main(x,n):
answer=1
mod=10**9+7
x_primes=prime_factorization(x)
for prime in x_primes:
power=0
factor=prime
while(factor<=n):
power+=n//factor
factor*=prime
answer*=pow(prime,power,mod)
answer%=mod
return answer
x,n=[int(c) for c in input().split()]
print(main(x,n))
``` | instruction | 0 | 41,344 | 22 | 82,688 |
Yes | output | 1 | 41,344 | 22 | 82,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's introduce some definitions that will be needed later.
Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.
Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:
* g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27),
* g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49).
Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:
* f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10,
* f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63.
You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.
Input
The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.
Output
Print the answer.
Examples
Input
10 2
Output
2
Input
20190929 1605
Output
363165664
Input
947 987654321987654321
Output
593574252
Note
In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.
In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).
In the third example, be careful about overflow issue.
Submitted Solution:
```
x, n = list(map(int, input().split()))
a = set()
if not x % 2:
a.add(2)
for i in range(3, int(x ** 0.5) + 1, 2):
if not x % i:
if all([i % j for j in a if j < (i + 1) ** 0.5]):
a.add(i)
if not len(a):
a.add(x)
a = sorted(a)
f, mod = 1, int(1e9 + 7)
for p in a:
m, c = n, 0
k, l = 1, p
while l <= m:
c += m // l
k += 1
l = p ** k
f *= pow(p, c, mod)
f = int(f)
print(f % mod)
``` | instruction | 0 | 41,345 | 22 | 82,690 |
No | output | 1 | 41,345 | 22 | 82,691 |
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