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>

using namespace std;

typedef long long ll;

const int INF = 1000000000;

int main()
{
  int N; scanf("%d", &N);
  vector<vector<pair<int, ll>>> adj(N + 5, vector<pair<int, ll>>());
  for (int i = 0; i < N - 1; ++i) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    adj[a].push_back(make_pair(b, c));
    adj[b].push_back(make_pair(a, c));
  }

  int Q, K;
  scanf("%d %d", &Q, &K);
  vector<ll> dist(N + 5, numeric_limits<ll>::max());
  dist[K] = 0;
  priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> que;
  que.push(make_pair(0, K));
  while (not que.empty()) {
    pair<ll, int> p = que.top(); que.pop();
    ll d = p.first;
    int u = p.second;
    if (d > dist[u]) { continue; }
    for (pair<ll, int> p : adj[u]) {
      int v = p.first;
      ll cost = p.second;
      if (d + cost < dist[v]) {
        dist[v] = d + cost;
        que.push(make_pair(dist[v], v));
      }
    }
  }
  for (int i = 0; i < Q; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%lld\n", dist[x] + dist[y]);
  }
  return 0;
}