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;
long long par[109][109];
void ini(long long n) {
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) par[i][j] = j;
  }
}
long long find(long long a, long long c) {
  if (par[c][a] == a) return a;
  return find(par[c][a], c);
}
void merge(long long a, long long b, long long c) {
  long long x = find(a, c), y = find(b, c);
  par[c][y] = par[c][x];
}
signed main() {
  long long n, m, i, j, k;
  cin >> n >> m;
  ini(105);
  for (i = 1; i <= m; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    merge(a, b, c);
  }
  cin >> k;
  while (k--) {
    long long x, y, ans = 0;
    cin >> x >> y;
    for (i = 1; i <= 100; i++) {
      long long a = find(x, i), b = find(y, i);
      if (a == b) ans++;
    }
    cout << ans << endl;
  }
}
