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<map<int, vector<int> > > v;
int n, m;
int dfs1(int parent, int curr, int find, int color, vector<int> visited) {
  int g = 0;
  visited[curr] = 1;
  vector<int> edges = (v[curr])[color];
  for (int i = 0; i < edges.size(); i++) {
    if (visited[edges[i]] == 0) {
      g += dfs1(curr, edges[i], find, color, visited);
      if (g > 0) return g;
    }
  }
  if (curr == find) g++;
  return g;
}
int dfs(int parent, int curr, int find) {
  int c = 0;
  for (auto it = v[curr].begin(); it != v[curr].end(); it++) {
    int color = it->first, g = 0;
    vector<int> visited(n, 0);
    visited[curr] = 1;
    vector<int> edges = it->second;
    for (int i = 0; i < edges.size(); i++) {
      int g = dfs1(curr, edges[i], find, color, visited);
      if (g != 0) {
        c += 1;
        break;
      }
    }
  }
  return c;
}
int main() {
  cin >> n >> m;
  v.resize(n);
  for (int i = 0; i < m; i++) {
    int k, l, co;
    cin >> k >> l >> co;
    k--;
    l--;
    ((v[k])[co]).push_back(l);
    ((v[l])[co]).push_back(k);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    int cal = dfs(-1, u, v);
    cout << cal << endl;
  }
  return 0;
}
