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 = 1e5 + 9;
unordered_map<int, int> par[MAX], sv[MAX];
int n, m, a, b, c, q;
int root(int v, int c) {
  return par[v][c] > 0 ? par[v][c] = root(par[v][c], c) : v;
}
void merge(int a, int b, int c) {
  if ((a = root(a, c)) == (b = root(b, c))) return;
  if (par[a][c] > par[b][c]) swap(a, b);
  par[a][c] += par[b][c] - 1, par[b][c] = a;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> a >> b >> c, merge(a, b, c);
  cin >> q;
  while (q--) {
    int ans = 0;
    cin >> a >> b;
    if (par[a].size() > par[b].size()) swap(a, b);
    if (sv[a].find(b) == sv[a].end()) {
      for (auto i : par[a])
        if (par[b].find(i.first) != par[b].end() &&
            root(b, i.first) == root(a, i.first))
          ans++;
      sv[a][b] = ans;
    }
    cout << sv[a][b] << '\n';
  }
}
