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 N = 1e9 + 7;
vector<pair<long long, long long>> adj[101];
map<pair<long long, long long>, long long> mp;
map<long long, long long> ans;
long long start, dest;
void dfs(long long node, long long col) {
  mp[{node, col}] = 1;
  if (node == dest) ans[col]++;
  for (auto child : adj[node]) {
    if (mp[{child.first, child.second}] == 0 &&
        (child.second == col || col == -1)) {
      dfs(child.first, child.second);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    long long u, v, c;
    cin >> u >> v >> c;
    adj[u].push_back({v, c});
    adj[v].push_back({u, c});
  }
  long long q;
  cin >> q;
  while (q--) {
    ans.clear();
    mp.clear();
    cin >> start >> dest;
    dfs(start, -1);
    long long temp = 0;
    for (auto x : ans) {
      if (x.second != 0) temp++;
    }
    cout << temp << "\n";
  }
}
