Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - ai, bi(1 ≤ ai < bi ≤ n) and ci(1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer- q(1 ≤ q ≤ 105), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - ui and vi(1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Examples

Input

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


Output

2
1
0


Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4


Output

1
1
1
1
2

Note

Let's consider the first sample. 

<image> The figure above shows the first sample. 

  * Vertex 1 and vertex 2 are connected by color 1 and 2. 
  * Vertex 3 and vertex 4 are connected by color 3. 
  * Vertex 1 and vertex 4 are not connected by any single color. 
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > adj[n];
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    u--;
    v--;
    c--;
    adj[u].push_back(make_pair(v, c));
    adj[v].push_back(make_pair(u, c));
  }
  int q;
  cin >> q;
  bool visited[n];
  while (q--) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    int count = 0;
    for (int i = 0; i < m; i++) {
      bool found = false;
      list<int> bfsqueue;
      for (int j = 0; j < n; j++) visited[j] = false;
      bfsqueue.push_back(u);
      visited[u] = true;
      while (!bfsqueue.empty()) {
        int f = *bfsqueue.begin();
        bfsqueue.erase(bfsqueue.begin());
        if (f == v) {
          found = true;
          break;
        }
        for (int j = 0; j < adj[f].size(); j++) {
          if (adj[f][j].second == i && !visited[adj[f][j].first]) {
            visited[adj[f][j].first] = true;
            bfsqueue.push_back(adj[f][j].first);
          }
        }
      }
      if (found) count++;
    }
    cout << count << endl;
  }
}
