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<vector<vector<int> > > g(101, vector<vector<int> >(101));
int n, m;
int vis[101][101];
void bfs(int s, int e) {
  set<int> col;
  queue<pair<int, pair<int, int> > > q;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < g[s][i].size(); ++j) {
      vis[s][g[s][i][j]] = 1;
      q.push(make_pair(g[s][i][j], make_pair(s, i)));
    }
  }
  while (q.size()) {
    int c = q.front().first;
    pair<int, int> p = q.front().second;
    q.pop();
    if (p.second == e) {
      col.insert(c);
      continue;
    }
    if (!vis[p.second][c]) {
      for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < g[p.second][i].size(); ++j) {
          if (g[p.second][i][j] == c) {
            if (!vis[i][c]) {
              q.push(make_pair(c, make_pair(p.second, i)));
            }
          }
        }
      }
      vis[p.second][c] = 1;
    }
  }
  cout << col.size() << endl;
}
int main() {
  cin >> n >> m;
  for (int i = 0, u, v, c; i < m; ++i) {
    cin >> u >> v >> c;
    g[u][v].push_back(c);
    g[v][u].push_back(c);
  }
  int t;
  cin >> t;
  while (t--) {
    memset(vis, 0, sizeof vis);
    int s, e;
    cin >> s >> e;
    bfs(s, e);
  }
}
