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;
bool reachable(int s, int d, vector<vector<int> > &ladj) {
  bool reached = false;
  vector<int> visited(ladj.size(), false);
  visited[s] = true;
  int current;
  stack<int> neighborhood;
  neighborhood.push(s);
  while (!reached && !neighborhood.empty()) {
    current = neighborhood.top();
    neighborhood.pop();
    for (int i = 0; i < ladj[current].size(); ++i) {
      if (ladj[current][i] == d) {
        reached = true;
      }
      if (!visited[ladj[current][i]]) {
        visited[ladj[current][i]] = true;
        neighborhood.push(ladj[current][i]);
      }
    }
  }
  return reached;
}
int main() {
  int i_nodes, i_edges;
  cin >> i_nodes >> i_edges;
  vector<vector<vector<int> > > edges_by_color(
      i_edges + 1, vector<vector<int> >(i_nodes + 1, vector<int>()));
  ;
  int u, v, c;
  for (int i = 0; i < i_edges; ++i) {
    cin >> u >> v >> c;
    edges_by_color[c][u].push_back(v);
    edges_by_color[c][v].push_back(u);
  }
  int queries;
  cin >> queries;
  int s, d;
  for (int i = 0; i < queries; ++i) {
    cin >> s >> d;
    int colors = 0;
    for (int j = 1; j < edges_by_color.size(); ++j) {
      if (reachable(d, s, edges_by_color[j])) {
        ++colors;
      }
    }
    cout << colors << endl;
  }
  return 0;
}
