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;
const long long mx = 1e6 + 9;
vector<pair<long long, long long> > adj[mx];
bool visited[mx];
set<long long> col;
long long flag;
void dfs(long long sor, long long des, long long cl) {
  if (sor == des) {
    flag = 1;
    return;
  }
  visited[sor] = true;
  for (long long i = 0; i < adj[sor].size(); i++) {
    long long v = adj[sor][i].first;
    long long c = adj[sor][i].second;
    if (visited[v] == false && cl == c) {
      dfs(v, des, c);
    }
  }
}
int main() {
  long long i, j, a, b, c, d, n, m, t, x, y, u, v, q;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> u >> v >> c;
    adj[u].push_back({v, c});
    adj[v].push_back({u, c});
    col.insert(c);
  }
  cin >> q;
  while (q--) {
    cin >> u >> v;
    long long cnt = 0;
    for (auto it : col) {
      c = it;
      flag = 0;
      memset(visited, 0, sizeof(visited));
      dfs(u, v, c);
      if (flag == 1) {
        cnt++;
      }
    }
    cout << cnt << endl;
  }
  return 0;
}
