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: Runtime Error)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MaxN = 1000;
typedef long long LL;

LL n, all, m;

int pre[MaxN + 5], last[MaxN + 5], other[MaxN + 5];
int seq[MaxN + 5];
LL dis[MaxN + 5], cost[MaxN + 5];
bool Inque[MaxN + 5];


void Build(int x, int y, LL w)
{
	pre[++all] = last[x];
	last[x] = all;
	other[all] = y;
	cost[all] = w;
}

void Bfs(int s)
{
	int head = 1, tail = 1, now, ed, dr;
	seq[tail] = s;
	for(int i = 1; i <= n; i++)
	{
		if(i == s) dis[i] = 0;
		else dis[i] = (LL)1 << 34;
	}
	Inque[s] = 1;
	while(head <= tail)
	{
		now = seq[head];
		ed = last[now];
		while(ed != -1)
		{
			dr = other[ed];
			if(dis[now] + cost[ed] < dis[dr])
			{
				dis[dr] = dis[now] + cost[ed];
				if(!Inque[dr])
				{
					Inque[dr] = 1;
					seq[++tail] = dr;
				}
			}
			ed = pre[ed];
		}
		Inque[now] = 0;
		head++;
	}
}


int main()
{
	all = -1;
	memset(last, -1, sizeof(last));
	scanf("%d", &n);
	for(int i = 1; i <= n - 1; i++)
	{
		int a, b;
		LL w;
		scanf("%d %d %lld", &a, &b, &w);
		Build(a, b, w);
		Build(b, a, w);
	}
	memset(seq, 0, sizeof(seq));
	memset(Inque, 0, sizeof(Inque));
	int a, b;
	scanf("%d %d", &a, &b);
	Bfs(b);
	for(int i = 1; i <= a; i++)
	{
		LL ans = 0;
		int x, y;
		scanf("%d %d", &x, &y);
		ans += dis[x];
		ans += dis[y];
		printf("%lld\n", ans);
	}
	return 0;
}