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 MAX_N = 111;
int cnt[MAX_N][MAX_N];
vector<int> l;
vector<pair<int, int> > gr[MAX_N];
int used[MAX_N];
inline void add(int v, int u, int c) { gr[v].push_back({u, c}); }
void dfs(int v, int c) {
  for (int i = 0; i < (int)l.size(); ++i) ++cnt[l[i]][v], ++cnt[v][l[i]];
  l.push_back(v);
  used[v] = c;
  for (int i = 0; i < (int)gr[v].size(); ++i) {
    if (gr[v][i].second != c) continue;
    int to = gr[v][i].first;
    if (used[to] == c) continue;
    dfs(to, c);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int v, u, c;
    cin >> v >> u >> c;
    add(v, u, c);
    add(u, v, c);
  }
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (used[j] < i) {
        l.clear();
        dfs(j, i);
      }
    }
  }
  int q;
  cin >> q;
  while (q-- > 0) {
    int v, u;
    cin >> v >> u;
    cout << cnt[v][u] << '\n';
  }
  return 0;
}
