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 <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define erep(i,a,b) for(int i=a;i<=(int)(b);++i)
#define per(i,a,b) for(int i=(b);i>(a);--i)
#define eper(i,a,b) for(int i=((int)(a));i>=b;--i)
#define pb push_back
#define mp make_pair
#define INF 1e9
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a, b)*b;}

const int MAX = 100005;
struct edge {int to; ll cost;};
vector<edge> g[MAX];
ll ans[MAX];

void dfs(int v, int p, ll d) {
    ans[v] = d;
    for (auto &e : g[v]) {
        if (e.to == p) continue;
        dfs(e.to, v, d + e.cost);
    }
}
int main() {
 ios::sync_with_stdio ( false );
 cin.tie ( 0 );
    int n;
    cin >> n;
    rep(i, 0, n-1) {
        int a, b, c;
        cin >> a >> b >> c;
        a--, b--;
        g[a].pb({b, c});
        g[b].pb({a, c});
    }
    int q, k;
    cin >> q >> k;
    k--;
    dfs(k, -1, 0);
    rep(i, 0, q) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        cout << ans[x] + ans[y] << endl;
    }
    return 0;
}