Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
struct Node {
  int num, pre;
  Node(int _num = 0, int _pre = 0) {
    num = _num;
    pre = _pre;
  }
};
int cnt = 0;
vector<bool> vi;
void dfs(vector<vector<vector<int> > > &arc, int n, int u, int v, int c,
         int pre) {
  vi[u] = true;
  int ii = 0, jj = 0;
  for (jj = 1; jj <= n; jj++) {
    if (jj != pre &&
        find(arc[u][jj].begin(), arc[u][jj].end(), c) != arc[u][jj].end() &&
        jj == v) {
      cnt++;
    }
    if (vi[jj] == false &&
        find(arc[u][jj].begin(), arc[u][jj].end(), c) != arc[u][jj].end()) {
      dfs(arc, n, jj, v, c, u);
    }
  }
  vi[u] = false;
}
int main() {
  int n = 0, m = 0;
  while (cin >> n >> m) {
    vector<int> v_tmp;
    vector<vector<vector<int> > > arc(n + 1,
                                      vector<vector<int> >(n + 1, v_tmp));
    int i = 0, j = 0;
    for (i = 1; i <= m; i++) {
      int a = 0, b = 0, c = 0;
      cin >> a >> b >> c;
      arc[a][b].push_back(c);
      arc[b][a].push_back(c);
    }
    int q = 0;
    cin >> q;
    int ii = 0, jj = 0, kk = 0;
    for (i = 0; i < q; i++) {
      int u = 0, v = 0;
      cin >> u >> v;
      if (u > v) {
        int tmp = u;
        u = v;
        v = tmp;
      }
      vi = vector<bool>(n + 1, false);
      cnt = 0;
      for (ii = 1; ii <= m; ii++) {
        dfs(arc, n, u, v, ii, -1);
      }
      cout << cnt << endl;
    }
  }
}
