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<pair<long long, long long> > v[500];
bool vis[500];
bool col[500];
bool f = 0;
void dfs(long long x, long long y, long long c) {
  vis[x] = true;
  if (x == y) f = 1;
  for (long long i = 0; i < v[x].size(); ++i) {
    if (!vis[v[x][i].first] && v[x][i].second == c) {
      dfs(v[x][i].first, y, c);
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  long long q, u, e, x, y, c, n, m;
  cin >> n >> m;
  while (m--) {
    cin >> x >> y >> c;
    v[x].push_back(make_pair(y, c));
    v[y].push_back(make_pair(x, c));
  }
  cin >> q;
  while (q--) {
    cin >> u >> e;
    long long ans = 0;
    memset(col, 0, sizeof(col));
    for (long long i = 0; i < v[u].size(); ++i) {
      f = 0;
      memset(vis, false, sizeof(vis));
      vis[u] = 1;
      if (!col[v[u][i].second]) {
        dfs(v[u][i].first, e, v[u][i].second);
        if (f) col[v[u][i].second] = 1;
        ans += f;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
