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 int inf = 1e9 + 5;
const long long linf = 1e18 + 5;
const int N = 1e5 + 5;
int n, m, x, y, t, root[N];
map<int, int> dp[N];
vector<pair<int, int> > v[N], comp[N];
int f(int x) {
  if (x == root[x]) return x;
  return root[x] = f(root[x]);
}
int main() {
  ios ::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y >> t;
    v[t].push_back(make_pair(x, y));
  }
  for (int i = 1; i <= m; i++) {
    for (__typeof((v[i]).begin()) it = (v[i]).begin(); it != (v[i]).end();
         it++) {
      root[it->first] = it->first;
      root[it->second] = it->second;
    }
    for (__typeof((v[i]).begin()) it = (v[i]).begin(); it != (v[i]).end();
         it++) {
      x = f(it->first);
      y = f(it->second);
      if (x == y) continue;
      root[y] = x;
    }
    for (__typeof((v[i]).begin()) it = (v[i]).begin(); it != (v[i]).end();
         it++) {
      if (comp[it->first].empty() or comp[it->first].back().first != i)
        comp[it->first].push_back(make_pair(i, f(it->first)));
      if (comp[it->second].empty() or comp[it->second].back().first != i)
        comp[it->second].push_back(make_pair(i, f(it->second)));
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    if (comp[x].size() > comp[y].size()) swap(x, y);
    if (dp[x].find(y) != dp[x].end()) {
      cout << dp[x][y] << '\n';
      continue;
    }
    int ans = 0;
    for (__typeof((comp[x]).begin()) it = (comp[x]).begin();
         it != (comp[x]).end(); it++)
      if (binary_search(comp[y].begin(), comp[y].end(), *it)) ans++;
    cout << ans << '\n';
    dp[x][y] = ans;
  }
  return 0;
}
