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;
vector<long long> adj[200];
vector<pair<long long, pair<long long, long long> > > p;
long long vis[210000];
void dfs(long long node) {
  vis[node] = 1;
  for (long long u = 0; u < adj[node].size(); u++) {
    if (!vis[adj[node][u]]) dfs(adj[node][u]);
  }
}
int main() {
  long long n, m, q, a, b, c, j;
  cin >> n >> m;
  for (long long i = 0; i < m; ++i) {
    cin >> a >> b >> c;
    p.push_back(make_pair(c, make_pair(a, b)));
  }
  sort(p.begin(), p.end());
  cin >> q;
  for (long long g = 0; g < q; ++g) {
    long long ans = 0;
    cin >> a >> b;
    long long x;
    for (long long i = 0; i < p.size();) {
      x = p[i].first;
      j = i;
      while (p[j].first == x) {
        adj[p[j].second.first].push_back(p[j].second.second);
        adj[p[j].second.second].push_back(p[j].second.first);
        j++;
      }
      dfs(a);
      if (vis[b]) ans++;
      i = j;
      for (long long k = 0; k < 200; k++) {
        vis[k] = 0;
        adj[k].clear();
      }
    }
    cout << ans << endl;
  }
  return 0;
}
