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;
const int INF = 2000000000;
const double pi = acos(-1.0);
int G[105][105][105], cc[105][105], vis[105];
int n, m;
void DFS(int c, int u, int ctr) {
  vis[u] = 1;
  cc[c][u] = ctr;
  for (int i = 1; i < n + 1; i++) {
    if (!vis[i] && (G[c][u][i] == 1)) {
      DFS(c, i, ctr);
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  while (cin >> n >> m) {
    memset(G, 0, sizeof(G));
    memset(cc, 0, sizeof(cc));
    int a, b, c;
    for (int i = 0; i < m; i++) {
      cin >> a >> b >> c;
      G[c][a][b] = 1;
      G[c][b][a] = 1;
    }
    for (int i = 1; i < m + 1; i++) {
      memset(vis, 0, sizeof(vis));
      int ct = 1;
      for (int j = 1; j < n + 1; j++) {
        if (!vis[j]) {
          DFS(i, j, ct);
          ct++;
        }
      }
    }
    int q;
    cin >> q;
    for (int i = 0; i < q; i++) {
      cin >> a >> b;
      int ans = 0;
      for (int j = 1; j < n + 1; j++)
        if ((cc[j][a] == cc[j][b]) && (cc[j][a] >= 1)) ans++;
      cout << ans << endl;
    }
  }
  return 0;
}
