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 <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <sstream>
#include <utility>
#define ll long long int
#define rep(i,x,y) for(int i=x;i<y;i++)
#define rel(i,x,y) for(int i=x-1;i>=y;i--)
#define all(x) x.begin(),x.end()
#define NMAX 100005
using namespace std;

ll ranges[NMAX] = {};
vector<pair<int,ll>> graph[NMAX]; //first is ngh. second is ranges.
bool used[NMAX] = {};


void dfs(int v,ll range){
	ranges[v] = range;
	used[v] = true;
	rep(i,0,graph[v].size()){
		if(!used[graph[v][i].first]){
			dfs(graph[v][i].first,range+graph[v][i].second);
		}
	}
}

int main(){
	int N;
	cin >> N;
	int tmp1,tmp2,tmp3;
	rep(i,0,N-1){
		cin >> tmp1 >> tmp2 >> tmp3;
		graph[tmp1].push_back(make_pair(tmp2,tmp3));
		graph[tmp2].push_back(make_pair(tmp1,tmp3));
	}
	int Q,K;
	int x,y;
	cin >> Q >> K;
	dfs(K,0);
	rep(i,0,Q){
		cin >> x >> y;
		cout << ranges[x] + ranges[y] << endl;
	}
}