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 <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;

struct edge
{
    int from,to;
    int cost;
};

long long d[100005];
bool vis[100005];

vector<edge> edges;
vector<int> v[100005];

void adde(int from,int to,int cost)
{
    edge e;
    e.from = from;
    e.to = to;
    e.cost = cost;

    edges.push_back(e);

    int m = edges.size() - 1;

    v[from].push_back(m);
}

void bfs(int s)
{
    vis[s] = 1;
    d[s] = 0;

    queue<int> q;

    q.push(s);

    while (!q.empty())
    {
        int now = q.front();q.pop();

        for (int i = 0;i < v[now].size();i++)
        {
            int id = v[now][i];

            edge tmp = edges[id];

            if (!vis[tmp.to])
            {
                vis[tmp.to] = 1;

                d[tmp.to] = d[now] + tmp.cost;

                q.push(tmp.to);
            }
        }
    }
}

int main()
{
    int n;

    scanf("%d",&n);

    for (int i = 1;i <= n - 1;i++)
    {
        int a,b,c;

        scanf("%d%d%d",&a,&b,&c);

        adde(a,b,c);
        adde(b,a,c);
    }

    int q,k;

    scanf("%d%d",&q,&k);

    bfs(k);

    for (int i = 0;i < q;i++)
    {
        int a,b;

        scanf("%d%d",&a,&b);

        printf("%lld\n",d[a] + d[b]);
    }

    return 0;
}