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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define INF 9e18
const int maxn = 4e5 + 10;
struct road{
	int v;
	int to;
	int next;
}r[maxn];
long long dis[maxn];
int head[maxn],cnt;
bool vis[maxn];
void inti(){
	memset(head, -1, sizeof(head));
	for(int i = 0; i < maxn; ++i){
		dis[i] = INF;
	}
	memset(vis, false, sizeof(vis));
	cnt = 0;
}
void build(long long s, long long e, long long w){
	r[cnt].v = w;
	r[cnt].to = e;
	r[cnt].next = head[s];
	head[s] = cnt++;
}
void spfa(long long s){
	queue<int>q;
	int end, i, top;
	q.push(s);
	vis[s] = true;
	dis[s] = 0;
	while(!q.empty()){
		top = q.front();
		q.pop();
		vis[top] = false;
		for(i = head[top]; i != -1; i = r[i].next){
			end = r[i].to;
			if(dis[end] > dis[top] + r[i].v){
				dis[end] = dis[top] + r[i].v;
				if(!vis[end]){
					q.push(end);
					vis[end] = true;
				}
			}
		}
	}
}
int main()
{
	long long n, s, e, w, q, k, i, a, b;
	scanf("%lld",&n);
	inti();
	for(i = 1; i < n; ++i){
		scanf("%lld %lld %lld", &s, &e, &w);
		build(s, e, w);
		build(e, s, w);
	}
	scanf("%lld %lld", &q, &k);
	spfa(k);
	for(i = 1; i <= q; ++i){
		scanf("%lld %lld", &a, &b);
		//cout << dis[a] << endl;
		printf("%lld\n",dis[a] + dis[b]);
	}
}