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;
long long int inf = pow(10, 9) + 7, ans;
void dfs(long long int x, long long int y,
         vector<vector<pair<long long int, long long int>>> &v, long long int c,
         vector<bool> &vis) {
  if (vis[x]) return;
  vis[x] = true;
  if (x == y) {
    ++ans;
    return;
  }
  for (long long int i = 0; i < v[x].size(); ++i) {
    if (v[x][i].second == c) dfs(v[x][i].first, y, v, c, vis);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, x, y, c;
  cin >> n >> m;
  vector<vector<pair<long long int, long long int>>> v(n + 1);
  for (long long int i = 0; i < m; ++i) {
    cin >> x >> y >> c;
    v[x].push_back({y, c});
    v[y].push_back({x, c});
  }
  long long int q;
  cin >> q;
  for (long long int i = 0; i < q; ++i) {
    cin >> x >> y;
    long long int z = 0;
    set<long long int> ass;
    for (long long int i = 0; i < v[x].size(); ++i) {
      ans = 0;
      vector<bool> vis(n + 1, false);
      if (ass.find(v[x][i].second) == ass.end())
        dfs(x, y, v, v[x][i].second, vis);
      ass.insert(v[x][i].second);
      z += ans;
    }
    cout << z << "\n";
  }
  return 0;
}
