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 f;
bool vis[105];
void dfs(vector<vector<pair<int, int> > > g, int t, int u, int v, int c) {
  vis[t] = true;
  for (int i = 0; i < g[t].size(); i++) {
    if (g[t][i].second == c) {
      if (vis[g[t][i].first] == false) {
        if (g[t][i].first == v)
          f = 1;
        else if (g[t][i].first == u) {
        } else
          dfs(g, g[t][i].first, u, v, c);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, i, j, q;
  cin >> n >> m;
  vector<vector<pair<int, int> > > g(n + 1);
  while (m--) {
    int u, v, c;
    cin >> u >> v >> c;
    g[u].push_back({v, c});
    g[v].push_back({u, c});
  }
  cin >> q;
  while (q--) {
    int u, v;
    f = 0;
    set<int> s;
    cin >> u >> v;
    for (i = 0; i < g[u].size(); i++) {
      f = 0;
      for (j = 0; j < 105; j++) vis[j] = false;
      if (g[u][i].first == v)
        s.insert(g[u][i].second);
      else {
        dfs(g, g[u][i].first, u, v, g[u][i].second);
        if (f == 1) s.insert(g[u][i].second);
      }
    }
    cout << s.size() << endl;
  }
  return 0;
}
