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 n, m, cn, q, ai, bi, ci, cnt;
vector<int> a[105][105];
bool v[105];
void f(int p, int q, int r) {
  int i;
  v[q] = true;
  if (q == r) {
    cnt++;
    return;
  }
  for (i = 0; i < a[p][q].size(); i++) {
    if (!v[a[p][q][i]]) f(p, a[p][q][i], r);
  }
}
int main() {
  int i, j, k;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    scanf("%d %d %d", &ai, &bi, &ci);
    a[ci][ai].push_back(bi);
    a[ci][bi].push_back(ai);
    if (ci > cn) cn = ci;
  }
  cin >> q;
  for (i = 0; i < q; i++) {
    scanf("%d %d", &ai, &bi);
    cnt = 0;
    for (j = 1; j <= cn; j++) {
      for (k = 1; k <= n; k++) {
        v[k] = false;
      }
      f(j, ai, bi);
    }
    cout << cnt << endl;
  }
}
