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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

typedef pair<long long, long long> pi;

long long n, a, b, c, q, k, x, y, dist[100001], p[100001][20], depth[100001];
vector<pi> adj[100001];

void dfs(long long node){
        for(vector<pi>::iterator it = adj[node].begin(); it != adj[node].end(); it++){
                if(it->first == 1 || depth[it->first] != 0 || dist[it->first] != 0) continue;
                dist[it->first] = dist[node]+it->second;
                p[it->first][0] = node;
                depth[it->first] = depth[node]+1;
                dfs(it->first);
        }
}

long long lca(long long x, long long y){
        if(depth[x] < depth[y]) swap(x, y);
        for(int k = 19; k >= 0; k--){
                if(p[x][k] != -1 && depth[p[x][k]] >= depth[y]) x = p[x][k];
        }
        if(x == y) return x;
        for(int k = 19; k >= 0; k--){
                if(p[x][k] != p[y][k]){
                        x = p[x][k];
                        y = p[y][k];
                }
        }
        return p[x][0];
}

long long distance(long long x, long long y){
        return dist[x]+dist[y]-2*dist[lca(x, y)];
}

int main(){
        cin >> n;
        for(int i = 0; i < n-1; i++){
                cin >> a >> b >> c;
                adj[a].push_back(pi(b, c));
                adj[b].push_back(pi(a, c));
        }
        memset(p, -1, sizeof(p));
        depth[1] = 0;
        dfs(1);
        for(int k = 1; k < 20; k++){
                for(int i = 1; i <= n; i++){
                        if(p[i][k-1] != -1) p[i][k] = p[p[i][k-1]][k-1];
                }
        }
        cin >> q >> k;
        for(int i = 0; i < q; i++){
                cin >> x >> y;
                cout << distance(x, k)+distance(k, y) << "\n";
        }
}