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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

const int maxn=1e6+5;

vector<pair<int,int> >e[maxn];//二维数组，第一维是点的序号，第二维是该点所连接的点的数量
//first是该点连接的点的序号，second是该点到所连接点的权值

int n,m,dis[maxn];

void init()
{
    for(int i=0;i<maxn;i++)e[i].clear();
    for(int i=0;i<maxn;i++)dis[i]=1e9;
}

void dij(int s)
{
    priority_queue<pair<int,int> >q;
    dis[s]=0;
    q.push(make_pair(-dis[s],s));
    while(!q.empty())
    {
        int now=q.top().second;//当前离起点最近的点（最短路已经确定的点）
        q.pop();
        for(int i=0;i<e[now].size();i++)
        {
            int v=e[now][i].first;//更新该点所连接的所有点
            if(dis[v]>dis[now]+e[now][i].second)
            {
                dis[v]=dis[now]+e[now][i].second;
                q.push(make_pair(-dis[v],v));//更新成功则入队
            }
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(0),cin.tie(0);
    while(cin>>n)
    {
        int q;
        init();
        for(int i=0;i<n-1;i++)
        {
            int x,y,z;
            cin>>x>>y>>z;
            e[x].push_back(make_pair(y,z));
            e[y].push_back(make_pair(x,z));
        }
        int s,t;
        cin>>q>>s;
        dij(s);
        while(q--)
        {
            int a,b;
            cin>>a>>b;
            cout<<dis[a]+dis[b]<<endl;
        }
    }
    return 0;
}