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 <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <set>
#include <iomanip>
#include <deque>
#include <stdio.h>
#include <random>
using namespace std;
 
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define RREP(i,n) for(int (i)=(int)(n)-1;i>=0;i--)
#define REMOVE(Itr,n) (Itr).erase(remove((Itr).begin(),(Itr).end(),n),(Itr).end())
typedef long long ll;

struct LCA {

    int N,LOG_N;
    vector< vector< int > > parent;
    vector< vector< int > > G;
    vector< int > depth;

    LCA (int n) : N(n), depth(n), G(n), LOG_N((int)(log2(n)+1)) {
        parent = vector< vector< int > >(LOG_N, vector<int>(N));
    }

    void add_edge(int a, int b) {
        G[a].push_back(b);
        G[b].push_back(a);
    }
    
    void init(int root) {
        dfs(root, -1, 0);
        for (int k = 0; k + 1 < LOG_N; k++) {
            for (int v = 0; v < N; v++) {
                if (parent[k][v] < 0) parent[k+1][v] = -1;
                else parent[k+1][v] = parent[k][parent[k][v]];
            }
        }
    }

    void dfs(int v, int p, int d) {
        parent[0][v] = p;
        depth[v] = d;
        for (int i = 0; i < G[v].size(); i++) {
            if (G[v][i] != p) dfs(G[v][i], v, d+1);
        }
    }

    int dist(int a, int b) {
        int par = getParent(a,b);
        return depth[a] + depth[b] - depth[par] * 2;
    }
    
    int getParent(int u, int v) {
        if (depth[u] > depth[v]) swap(u,v);
        for (int k = 0; k < LOG_N; k++) {
            if (((depth[v] - depth[u]) >> k) & 1) v = parent[k][v];
        }
        if (u == v) return u;
        for (int k = LOG_N - 1; k >= 0; k--) {
            if(parent[k][u] != parent[k][v]){
                u = parent[k][u];
                v = parent[k][v];
            }
        }
        return parent[0][u];
    }

};

template <class T> struct Dijkstra {

    struct Edge{
        int to;
        T cost;
        Edge(int _to, T _cost) : to(_to), cost(_cost) {}
    };
    
    const T INF_T = numeric_limits<T>::max() / 2;
    vector< vector< Edge > > G;
    vector< T > d;
    Dijkstra (int n) : G(n), d(n) {}

    void add_directed_edge(int a, int b, T c) {
        G[a].push_back(Edge(b,c));
    }
    
    void add_undirected_edge(int a, int b, T c) {
        G[a].push_back(Edge(b,c));
        G[b].push_back(Edge(a,c));
    }

    void init(int s) {
        priority_queue< pair<T,int>,vector< pair<T,int> >, greater< pair<T,int> > > que;
        for (int i = 0; i < d.size(); i++) d[i] = INF_T;
        d[s] = 0;
        que.push(make_pair(0,s));
        while (!que.empty()){
            pair<T,int> p = que.top();
            que.pop();
            int v = p.second;
            if (d[v] < p.first) continue;
            for (int i = 0; i < G[v].size(); i++) {
                Edge e = G[v][i];
                if (d[e.to] > d[v] + e.cost) {
                    d[e.to] = d[v] + e.cost;
                    que.push(make_pair(d[e.to],e.to));
                }
            }
        }
    }
    
    T dist(int a) {
        return d[a];
    }
    
}; 


int main() {

    LCA inst(100010);
    Dijkstra<ll> inst1(100010);

    int N; cin >> N;
    REP(i,N-1) {
        int a,b;
        ll c;
        cin >> a >> b >> c;
        inst.add_edge(a,b);
        inst1.add_undirected_edge(a,b,c);
    }

    inst.init(1);

    int Q,K; cin >> Q >> K;    
    inst1.init(K);
    REP(i,Q) {
        int a,b;
        cin >> a >> b;
        if(inst.dist(a,b) == inst.dist(a,K) + inst.dist(b,K)) {
            cout << inst1.dist(a) + inst1.dist(b) << endl;
        }else {
            cout << inst1.dist(a) + inst1.dist(b) << endl;
        }
    }

    return 0;
}