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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

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

typedef struct ADJ_Tag{
	int v;
	int c;
	struct ADJ_Tag *next;
} ADJ;

int n, q, k;
ADJ *graph[100001];
long long int depth[100001];
int flags[100001];

void new_ADJ( ADJ **p, int b, int c ){
	*p = (ADJ *)malloc(sizeof(ADJ));
	if(*p == NULL){
		exit(1);
	}
	(*p)->v = b;
	(*p)->c = c;
	(*p)->next = NULL;
	return;
}

void add( int a, ADJ *new ){
	if( graph[a] == NULL ){
		graph[a] = new;
		
	}else{
		ADJ *p = graph[a];
		while( p->next != NULL ){
			p = p->next;
		}
		p->next = new;	
	}
	return;	
}

void dfs( int v, long long int d ){
	depth[v] = d;
	flags[v] = 1;
	ADJ *p = graph[v];
	while( p != NULL ){
		if( flags[p->v] == 0 ){
			dfs( p->v, d + (long long int)(p->c) );
		}
		p = p->next;
	}
	return;
}

int main(){
	
	int i;
	for( i = 0; i < 100001; i++ ){
		graph[i] = NULL;
		flags[i] = 0;
	}
	
	scanf("%d", &n);
	for( i = 1; i < n; i++ ){
		int a, b, c;
		scanf( "%d %d %d", &a, &b, &c );
		ADJ *new;
		new_ADJ( &new, b, c );
		add( a, new );
		new_ADJ( &new, a, c );
		add( b, new );
	}

	scanf( "%d %d", &q, &k );
	dfs( k, 0 );
	
	int x, y;
	for( i = 1; i <= q; i++ ){
		scanf( "%d %d", &x, &y );
		printf( "%lld\n", depth[x] + depth[y] );
	}
	
	return 0;
	
}