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# 1253_F. Cheap Robot

## Problem Description
You're given a simple, undirected, connected, weighted graph with n nodes and m edges.

Nodes are numbered from 1 to n. There are exactly k centrals (recharge points), which are nodes 1, 2, …, k.

We consider a robot moving into this graph, with a battery of capacity c, not fixed by the constructor yet. At any time, the battery contains an integer amount x of energy between 0 and c inclusive.

Traversing an edge of weight w_i is possible only if x β‰₯ w_i, and costs w_i energy points (x := x - w_i).

Moreover, when the robot reaches a central, its battery is entirely recharged (x := c).

You're given q independent missions, the i-th mission requires to move the robot from central a_i to central b_i.

For each mission, you should tell the minimum capacity required to acheive it.

Input

The first line contains four integers n, m, k and q (2 ≀ k ≀ n ≀ 10^5 and 1 ≀ m, q ≀ 3 β‹… 10^5).

The i-th of the next m lines contains three integers u_i, v_i and w_i (1 ≀ u_i, v_i ≀ n, u_i β‰  v_i, 1 ≀ w_i ≀ 10^9), that mean that there's an edge between nodes u and v, with a weight w_i.

It is guaranteed that the given graph is simple (there is no self-loop, and there is at most one edge between every pair of nodes) and connected.

The i-th of the next q lines contains two integers a_i and b_i (1 ≀ a_i, b_i ≀ k, a_i β‰  b_i).

Output

You have to output q lines, where the i-th line contains a single integer : the minimum capacity required to acheive the i-th mission.

Examples

Input


10 9 3 1
10 9 11
9 2 37
2 4 4
4 1 8
1 5 2
5 7 3
7 3 2
3 8 4
8 6 13
2 3


Output


12


Input


9 11 3 2
1 3 99
1 4 5
4 5 3
5 6 3
6 4 11
6 7 21
7 2 6
7 8 4
8 9 3
9 2 57
9 3 2
3 1
2 3


Output


38
15

Note

In the first example, the graph is the chain 10 - 9 - 2^C - 4 - 1^C - 5 - 7 - 3^C - 8 - 6, where centrals are nodes 1, 2 and 3.

For the mission (2, 3), there is only one simple path possible. Here is a simulation of this mission when the capacity is 12.

  * The robot begins on the node 2, with c = 12 energy points. 
  * The robot uses an edge of weight 4.
  * The robot reaches the node 4, with 12 - 4 = 8 energy points. 
  * The robot uses an edge of weight 8.
  * The robot reaches the node 1 with 8 - 8 = 0 energy points. 
  * The robot is on a central, so its battery is recharged. He has now c = 12 energy points. 
  * The robot uses an edge of weight 2.
  * The robot is on the node 5, with 12 - 2 = 10 energy points. 
  * The robot uses an edge of weight 3.
  * The robot is on the node 7, with 10 - 3 = 7 energy points. 
  * The robot uses an edge of weight 2.
  * The robot is on the node 3, with 7 - 2 = 5 energy points. 
  * The robot is on a central, so its battery is recharged. He has now c = 12 energy points. 
  * End of the simulation. 



Note that if value of c was lower than 12, we would have less than 8 energy points on node 4, and we would be unable to use the edge 4 ↔ 1 of weight 8. Hence 12 is the minimum capacity required to acheive the mission.

β€”

The graph of the second example is described here (centrals are red nodes):

<image>

The robot can acheive the mission (3, 1) with a battery of capacity c = 38, using the path 3 β†’ 9 β†’ 8 β†’ 7 β†’ 2 β†’ 7 β†’ 6 β†’ 5 β†’ 4 β†’ 1

The robot can acheive the mission (2, 3) with a battery of capacity c = 15, using the path 2 β†’ 7 β†’ 8 β†’ 9 β†’ 3

## Contest Information
- **Contest ID**: 1253
- **Problem Index**: F
- **Points**: 2750.0
- **Rating**: 2500
- **Tags**: binary search, dsu, graphs, shortest paths, trees
- **Time Limit**: {'seconds': 3, 'nanos': 0} seconds
- **Memory Limit**: 512000000 bytes

## Task
Solve this competitive programming problem. Provide a complete solution that handles all the given constraints and edge cases.