Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
vector<long long int> adj[101];
set<pair<long long int, long long int>> color[101];
long long int visited[101];
long long int x;
void bfs(long long int a, long long int b, long long int c) {
  queue<long long int> q;
  q.push(a);
  while (!q.empty()) {
    long long int p = q.front();
    q.pop();
    if (p == b) {
      x++;
      visited[p] = 1;
    }
    if (visited[p]) continue;
    visited[p] = 1;
    for (long long int i = 0; i < adj[p].size(); i++) {
      if (color[c].find({p, adj[p][i]}) != color[c].end() &&
          visited[adj[p][i]] == 0) {
        q.push(adj[p][i]);
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k, l, m, n, p, t, a, b, c, q;
  cin >> n >> m;
  set<pair<long long int, long long int>> s;
  for (i = 0; i < m; i++) {
    cin >> a >> b >> c;
    if (s.find({a, b}) == s.end()) {
      s.insert({a, b});
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    color[c].insert({a, b});
    color[c].insert({b, a});
  }
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> a >> b;
    x = 0;
    for (j = 1; j <= m; j++) {
      bfs(a, b, j);
      memset(visited, 0, sizeof(visited));
    }
    cout << x << "\n";
  }
}
