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;
using pr = pair<int, int>;
using vp = vector<pr>;
using vvp = vector<vp>;
using vi = vector<int>;
int N;
void dfs(vvp& g, vi& vis, int& fl, int fin, int col, int nd) {
  if (nd == fin) {
    fl = 1;
    return;
  }
  if (fl) return;
  vis[nd] = 1;
  for (auto& pr : g[nd]) {
    if (pr.second == col)
      if (!vis[pr.first]) dfs(g, vis, fl, fin, col, pr.first);
  }
}
int get_num_colors(vvp& g, int u, int v) {
  int ans = 0;
  for (auto& pr : g[u]) {
    vi vis(N + 1);
    int fl = 0;
    dfs(g, vis, fl, v, pr.second, u);
    ans += fl;
  }
  return ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  N = n;
  vvp g(n + 1);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[a].push_back({b, c});
    g[b].push_back({a, c});
  }
  int q;
  cin >> q;
  map<pr, int> mp;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    if (mp.count({u, v})) {
      int ans = mp[{u, v}];
      cout << ans << endl;
    } else {
      int ans = get_num_colors(g, u, v);
      mp[{u, v}] = ans;
      mp[{v, u}] = ans;
      cout << ans << endl;
    }
  }
  return 0;
}
