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;
bool visited[101] = {0};
vector<pair<int, int>> v[101];
bool dfs(int s, int d, int c) {
  visited[s] = true;
  if (s == d) return 1;
  for (int i = 0; i < v[s].size(); i++) {
    pair<int, int> p = v[s][i];
    if (p.second == c && !visited[p.first]) {
      if (dfs(p.first, d, c)) return 1;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tests = 1;
  while (tests-- > 0) {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
      int a, b, c;
      cin >> a >> b >> c;
      v[a].push_back({b, c});
      v[b].push_back({a, c});
    }
    int q;
    cin >> q;
    while (q-- > 0) {
      int p, q, ans = 0;
      cin >> p >> q;
      for (int i = 1; i <= m; i++) {
        memset(visited, 0, sizeof(visited));
        if (dfs(p, q, i)) ans++;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
