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>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
using namespace std;

const long long INF = (1LL << 60);

class Edge{
    public:
        long long to, cost;
        Edge(long long to, long long cost) : to(to) ,cost(cost) {}
};

class Node{
    public:
        long long dis;
        bool isUsed;
        Node(){
            this->dis = INF;
            this->isUsed = 0;
        }
};

typedef vector<vector<Edge>> AdjList;

vector<Node> dijkstra(AdjList g, long long start, long long n){
    vector<Node> node(n);
    priority_queue<long long, vector<pair<long long, long long>>, greater<pair<long long, long long>>> q;

    q.push(make_pair(0, start));
    node[start].dis = 0;

    pair<long long, long long> u;
    while(not q.empty()){
        u = q.top(); q.pop();
        long long current = u.second;
        node[current].isUsed = 1;

        rep(i,g[current].size()){
            long long next = g[current][i].to;
            if(node[next].isUsed == 0){
                if(node[next].dis > node[current].dis + g[current][i].cost){
                    node[next].dis = node[current].dis + g[current][i].cost;
                    q.push(make_pair(node[next].dis, next));
                }
            }
        }
    }
    return node;
}

int main(){
    long long n;
    cin >> n;
    AdjList g(n);

    rep(i,n - 1){
        long long a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        g[a].emplace_back(Edge{b,c});
        g[b].emplace_back(Edge{a,c});
    }

    long long q, k;
    cin >> q >> k;
    k--;

    vector<Node> node = dijkstra(g, k, n);
    rep(i,q){
        long long a, b;
        cin >> a >> b;
        a--; b--;
        cout << node[a].dis + node[b].dis << endl;
    }

}