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 (Clang 3.8.0) (judge verdict: Memory Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

typedef long long ll;
typedef struct {
	ll cost;
	int to;
} Edge;

ll n;
ll q, k;
ll min_cost[100005];
int visited[100005];
Edge *graph[100005];
int edge_num[100005];

int st[100005];
int sp;

void dfs(ll k);

int main()
{
	scanf("%lld", &n);

	for (ll i = 1; i <= n; i++) {
		graph[i] = malloc(sizeof(Edge) * n);
	}

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

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

		graph[a][edge_num[a]] = (Edge){.cost = c, .to = b};
		graph[b][edge_num[b]] = (Edge){.cost = c, .to = a};

		edge_num[a]++;
		edge_num[b]++;

	}

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

	dfs(k);

	for (ll i = 0; i < q; i++) {
		ll x, y;

		scanf("%lld %lld", &x, &y);

		printf("%lld\n", min_cost[x] + min_cost[y]);
	}


	return 0;
}

void dfs(ll k)
{
	visited[k] = 1;
	st[0] = k;
	sp = 0;

	while(sp >= 0) {
		ll pos = st[sp]; sp--;

		for (ll i = 0; i < edge_num[pos]; i++) {
			ll npos = graph[pos][i].to;
			ll ncos = graph[pos][i].cost + min_cost[pos];

			if (visited[npos] == 0) {
				visited[npos] = 1;
				min_cost[npos] = ncos;

				sp++;
				st[sp] = npos;
			}
		}
	}

	return;
}