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: Accepted)

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

typedef struct{
    long long st;
    long long fi;
    long long kr;
}rs;

typedef struct{
    long long st;
    long long kz;
}mkj;

int sortfnc(const void *a,const void *b){
if(((rs*)a)->st == ((rs*)b)->st){return 0;}
if(((rs*)a)->st < ((rs*)b)->st){return -1;}
return 1;
}

void makemkj(rs g[],mkj x[],long long n){
    long long i,ms=0,nst=g[0].st;
    for(i=1;i<n;i++){
        if(g[i].st!=g[i-1].st){
            x[nst].kz=i-ms;
            x[nst].st=ms;
            nst=g[i].st;ms=i;
        }
    }
    x[nst].kz=n-ms;
    x[nst].st=ms;
}

void calcdif(rs g[],mkj x[],long long nc,long long ndif,long long dif[]){
    long long i;
    for(i=x[nc].st;i<x[nc].st+x[nc].kz;i++){
        if(dif[g[i].fi] != -1){continue;}
        dif[g[i].fi]=ndif+g[i].kr;
        calcdif(g,x,g[i].fi,dif[g[i].fi],dif);
    }
}

int main(void){
    long long n,a[131072],b[131072],c[131072],i;
    long long q,k;
    long long z[131072],y[131072];
    long long dif[131072];
    rs g[262144];
    mkj x[131072];
    scanf("%lld",&n);
    for(i=0;i<n-1;i++){
        scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);
        g[2*i].st=a[i];g[2*i].fi=b[i];g[2*i].kr=c[i];
        g[2*i+1].st=b[i];g[2*i+1].fi=a[i];g[2*i+1].kr=c[i];
    }
    qsort(g,2*n,sizeof(rs),sortfnc);
    makemkj(g,x,2*n);
    scanf("%lld%lld",&q,&k);
    for(i=0;i<=131071;i++){dif[i]=-1;}
    dif[k]=0;
    calcdif(g,x,k,0,dif);
    for(i=1;i<=q;i++){
        scanf("%lld%lld",&z[i],&y[i]);
        printf("%lld\n",dif[z[i]]+dif[y[i]]);
    }
    return 0;
}