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;
int i, j, k, n, m, x, y, T, num, w, c, q, sum;
bool g[105][105][105], color[105];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &c);
    color[c] = 1;
    g[c][x][y] = 1;
    g[c][y][x] = 1;
  }
  for (c = 1; c <= m; c++) {
    if (!color[c]) continue;
    for (k = 1; k <= n; k++) {
      for (i = 1; i <= n; i++) {
        if (i == k) continue;
        for (j = 1; j <= n; j++) {
          if (k == j || i == j) continue;
          g[c][i][j] = g[c][i][j] | (g[c][i][k] & g[c][k][j]);
        }
      }
    }
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &x, &y);
    sum = 0;
    for (j = 1; j <= m; j++) {
      if (g[j][x][y]) sum++;
    }
    printf("%d\n", sum);
  }
  return 0;
}
