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;
int n, m, a, b, c, k, pr, ans, vis[109];
vector<pair<int, int> > v[109];
void dfs(int a, int b) {
  for (auto i : v[a]) {
    if (i.first == b and i.second == pr) {
      ans++;
    }
    if (i.second == pr and i.first != b and !vis[i.first])
      vis[a] = 1, dfs(i.first, b);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    v[b].push_back({a, c});
  }
  cin >> k;
  for (int i = 1; i <= k; i++) {
    memset(vis, 0, sizeof(vis));
    cin >> a >> b;
    ans = 0;
    vis[a] = 1;
    for (auto j : v[a]) {
      if (j.first == b) ans++;
      pr = j.second;
      if (j.first != b) dfs(j.first, b);
    }
    cout << ans << endl;
  }
}
