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.

Examples

Input

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5


Output

3
2
4


Input

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


Output

5
14
22


Input

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


Output

17000000000
#include <bits/stdc++.h>

using namespace std;

const int N = 100005;

long long d[N];
vector<pair<int, int>> g[N];

void dfs(int u, int f, long long w) {
  d[u] = w;
  for (auto &go : g[u]) {
    if (go.first != f) {
      dfs(go.first, u, w + go.second);
    }
  }
}

int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  int q, k;
  scanf("%d %d", &q, &k);
  dfs(k, k, 0);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    cout << d[u] + d[v] << endl;
  }
  return 0;
}
