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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

const ll INF = 1LL<<60;

struct edge {
    int to;
    int cost;
    edge(int t, int w) : to(t), cost(w) { }
};

int n, q, k;
vector<vector<edge>> tree;

void solve(int x, int y) {   
    priority_queue<P, vector<P>, greater<P>> q;
    vector<ll> dist(n, INF);
    q.push(P(0, x));
    dist[x] = 0;
    
    while (!q.empty()) {
        auto p = q.top();
        q.pop();
        int v = p.second;
        if (dist[v] < p.first) continue;
        for (int i = 0; i < tree[v].size(); ++i) {
            int nv = tree[v][i].to, weight = tree[v][i].cost;
            if (dist[nv] <= dist[v] + weight) continue;
            dist[nv] = dist[v] + weight;
            q.push(P(dist[nv], nv));
        }
    }

    ll res = dist[k];

    dist = vector<ll>(n, INF);
    q.push(P(0, k));
    dist[k] = 0;
    
    while (!q.empty()) {
        auto p = q.top();
        q.pop();
        int v = p.second;
        if (dist[v] < p.first) continue;
        for (int i = 0; i < tree[v].size(); ++i) {
            int nv = tree[v][i].to, weight = tree[v][i].cost;
            if (dist[nv] <= dist[v] + weight) continue;
            dist[nv] = dist[v] + weight;
            q.push(P(dist[nv], nv));
        }
    }
    
    res += dist[y];
    cout << res << endl;
    return;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    cin >> n;
    tree.resize(n);
    rep(i, n-1) {
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        tree[a].emplace_back(b, c);
        tree[b].emplace_back(a, c);
    }
    cin >> q >> k;
    k--;
    rep(i, q) {
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        solve(x, y);
    }
}