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<cstdio>
#include<algorithm>
#include<functional>
#define F1 first.first
#define F2 first.second
#define F3 second
#define MM 10000000000000000
using namespace std;
void func(int);
typedef pair<int,int> P1;
typedef pair<P1,int> P2;
P2 p[200000];
int st[100002];
long long leng[100001];
int main(void)
{
	int n,i,j,a,b,c,q,k,n2,x,y;

	scanf("%d",&n);
	for(i=0;i<n-1;i++)	{
		scanf("%d %d %d",&a,&b,&c);
		p[i].F1=a;	p[i].F2=b;	p[i].F3=c;
		p[i+n-1].F1=b;
		p[i+n-1].F2=a;
		p[i+n-1].F3=c;
	}
	n2=(n-1)*2;
	sort(p,p+n2);
	st[1]=0;
	for(i=1;i<n2;i++)	{
		if(p[i].F1!=p[i-1].F1)	{
			a=p[i].F1;
			st[a]=i;
		}
	}
	st[n+1]=n2;
//	for(i=0;i<n2;i++)	{
//		printf(" %d %d %d\n",p[i].F1,p[i].F2,p[i].F3);
//	}
//	for(i=1;i<=n;i++)	printf("st=%d\n",st[i]);
	scanf("%d %d",&q,&k);
	for(i=1;i<=n;i++)	leng[i]=MM;
	leng[k]=0;
	func(k);
	for(i=0;i<q;i++)	{
		scanf("%d %d",&x,&y);
		printf("%lld\n",leng[x]+leng[y]);
	}
	return 0;
}
void func(int a)
{
	int i,p1,p2,xx,l2;
	p1=st[a];	p2=st[a+1];
	for(i=p1;i<p2;i++)	{
		xx=p[i].F2;
		l2=p[i].F3;
		if(leng[xx]>leng[a]+l2)	{
			leng[xx]=leng[a]+l2;
			func(xx);
		}
	}
}