Problem:
Score : 400 points

Problem Statement

You are given a tree with N vertices.

Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N-1 edges, where N is the number of its vertices.

The i-th edge (1≤i≤N-1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

3≤N≤10^5

1≤a_i,b_i≤N (1≤i≤N-1)

1≤c_i≤10^9 (1≤i≤N-1)

The given graph is a tree.

1≤Q≤10^5

1≤K≤N

1≤x_j,y_j≤N (1≤j≤Q)

x_j≠y_j (1≤j≤Q)

x_j≠K,y_j≠K (1≤j≤Q)

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
:
a_{N-1} b_{N-1} c_{N-1}
Q K
x_1 y_1
:
x_{Q} y_{Q}

Output

Print the responses to the queries in Q lines.

In the j-th line j(1≤j≤Q), print the response to the j-th query.

Sample Input 1

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3

Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2

Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.

Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int n,par[100020],a[100020],b[100020],q,k;
long long dis[100020],c[100020];

int find (int x){
    if (par[x]==x){
        return x;
    } else {
        int p=find(par[x]);
        dis[x]=dis[par[x]]+dis[x];
        par[x]=p;
        return par[x];
    }
}

void unite (int x, int y, long long d){
    int px=find(x);
    int py=find(y);
    if (x!=y){
        if (px==k){
            par[py]=k;
            dis[py]=dis[x]+d-dis[y];
        } else if (py==k){
            par[px]=k;
            dis[px]=dis[y]+d+dis[x];
        } else {
            par[px]=py;
            dis[px]=dis[y]+d+dis[x];
        }
    }
}


int main(void){
    scanf("%d",&n);
    for (int i=0; i<n; i++){
        par[i]=i;
    }
    for (int i=0; i<n-1; i++){
        scanf("%d%d%lld",&a[i],&b[i],&c[i]);
        a[i]--;
        b[i]--;
    }
    scanf("%d%d",&q,&k);
    k--;
    for (int i=0; i<n-1; i++){
        unite(a[i],b[i],c[i]);
    }
    for (int i=0; i<q; i++){
        int x,y;
        scanf("%d%d",&x,&y);
        printf("%lld\n",dis[x-1]+dis[y-1]);
    }
	return 0 ;
}