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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 ≥ 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 ⋅ 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 ≠ 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 ≠ 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

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.

—

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

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

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