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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

const int MAX_INDEX = pow(10, 5) + 1;
const ll NO_ROUTE = 1234560000000L;

struct Edge {
    int to;
    ll distance;
};
struct Data {
    int from;
    int index;
    ll distanceFromK;
};
//
int N;
//経路
vector<vector<Edge>> edgeMap(10001);
//Kからの距離
ll minD[10001];

//全都市のKからの距離を計算
//K,およびその先から伸びるすべての辺を探索。最大でも10000回ですむはず
void calcFromK(int k) {
    list<Data> queue;
    queue.push_back({-1, k, 0});

    while (queue.size() > 0) {
        Data d = queue.front();
        queue.pop_front();
        //都市 d.curr から伸びている辺を全部あさる。ただし来た道は除外
        auto arr = edgeMap[d.index];
        minD[d.index] = d.distanceFromK;
        for (auto e: arr) {
            if (e.to != d.from) {
                queue.push_back({d.index, e.to, e.distance + d.distanceFromK});
            }
        }
    }
}
//x -> k -> yな距離を計算
ll solve(int x, int y, int k) {
    return minD[x] + minD[y];
}

int main()
{
    cin >> N;
    rep(i, N - 1) {
        int a, b, c;
        cin >> a >> b >> c;
        edgeMap[a].push_back({b, c});
        edgeMap[b].push_back({a, c});
    }
    int Q, K;
    cin >> Q >> K;
    vector<ll> ans;
    calcFromK(K);
    rep(i, Q) {
        int x, y;
        cin >> x >> y;
        ans.push_back(solve(x, y, K));
    }

    for (auto a : ans) {
        cout << a << endl;
    }
    return 0;
}