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<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<deque>
#include<map>

#define rep(i, n) for(int i = 0;i < (n);i++)
#define rrep(i, n) for(int i = (n);i >= 0;i--)
#define FOR(i, m, n) for(int i = (m);i < (n);i++)

using namespace std;
using ll = long long;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1;} return 0;}
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1;} return 0;}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};


//////////////////////////////////////////////////

struct Edge {
    int to;
    ll cost;
};

int main() {
    int n; cin >> n;
    vector<vector<Edge>> g(n);
    for (int i = 0; i < n-1; i++)
    {
        int a, b; cin >> a >> b;
        a--; b--;
        ll c; cin >> c;

        g[a].push_back(Edge{b, c});
        g[b].push_back(Edge{a, c});
    }

    int q, k; cin >> q >> k;
    k--;
    vector<pair<int, int>> xy(n);
    for (int i = 0; i < q; i++)
    {
        int x, y;
        cin >> x >> y;
        x--; y--;
        xy[i] = make_pair(x, y);
    }


    // dijk
    // sからの距離を返す
    int s = k;
    const ll inf = 1e14;
    vector<ll> dp(n, inf);
    dp[s] = 0;
    vector<bool> visited(n, false);
    
    auto dfs = [&](auto &&dfs, int now, ll len) -> void {
        if (visited[now]) return;
        visited[now] = true;
        dp[now] = len;

        for (int i = 0; i < g[now].size(); i++)
        {
            int next = g[now][i].to;
            ll cost = g[now][i].cost;
            dfs(dfs, next, dp[now]+cost);
        }
        
    };
    
    dfs(dfs, s, 0);

    for (int i = 0; i < q; i++)
    {
        cout << dp[xy[i].first] + dp[xy[i].second] << endl;
    }
    
    
    return 0;
}