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 (Clang 3.8.0) (judge verdict: Accepted)

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

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

// array pointer = *a, num of element = n, key key
int lower_bound(int *a, int n, int key){
    int ng, mid, ok;
    ng = -1, ok = n-1;
    while (abs(ok - ng) > 1){
        mid = (ok + ng) / 2;
        if (key <= a[mid]){
            ok = mid;
        }else{
            ng = mid;
        }
    }
    if (a[ok] >= key)return ok;
    return n;
}

//greatest common divisor
unsigned long gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}

unsigned long lcm(unsigned long x, unsigned long y){
    unsigned long g = gcd(x, y);
    return x*y/g;

}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

/*
int struct_ascending_compare(const void *p, const void *q) {
return ((struct_name*)p)->member_name - ((struct_name*)q)->member_name;
}*/

/*unsigned long long pascal[100][100] = {0};
void make_pascal(void){
    for (int i = 0; i < 100; i++){
        pascal[i][0] = 1;
    }
    pascal[1][1] = 1;
    for (int i = 2; i < 100; i++){
        for (int j = 1; j < 100; j++){
           pascal[i][j] = (pascal[i-1][j-1]+pascal[i-1][j]) % mod;
        }
    }
}*/
long long mod = 1000000007;

//x ^ n
long long mod_pow(long long x, long long n){
    long long res = 1;
    for(int i = 0;i < 60; i++){
        if(n >> i & 1) res = res * x % mod;
        x = x * x % mod;
    }
    return res;
}

int n;

int to[200005];
int head[100005];
int next[200005];
int cost[200005];
long long d[100005];
const long long INF = 1000000000000000;
int x[100005];
int y[100005];
long long ans[100005];

int dfs(int crt){
    int edge = head[crt];
    while (edge != -1){
        if (d[to[edge]] > d[crt] + cost[edge]){
            d[to[edge]] = d[crt] + cost[edge];
            dfs(to[edge]);
        }
        edge = next[edge];
    }
    return 0;
}

int main(void){
    scanf("%d", &n);
    int a, b, c;
    for (int i = 1; i <= n; i++){
        head[i] = -1;
        d[i] = INF;
    }
    for (int i = 0; i < (n-1)*2; i += 2){
        scanf("%d %d %d", &a, &b, &c);
        next[i] = head[a];
        head[a] = i;
        to[i] = b;
        cost[i] = c;

        next[i+1] = head[b];
        head[b] = i+1;
        to[i+1] = a;
        cost[i+1] = c;
    }
    int q, k;
    scanf("%d %d", &q, &k);
    d[k] = 0;
    dfs(k);
    for (int i = 0; i < q; i++){
        scanf("%d %d", &x[i], &y[i]);
        ans[i] = d[x[i]]+d[y[i]];
    }

    for (int i = 0; i < q; i++){
        printf("%lld\n", ans[i]);
    }


    return 0;
}