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;
bool comp(int a, int b) { return (a > b); }
int ans[102][102] = {};
int parent(int color, int x) {
  return ans[color][x] == x ? x : ans[color][x] = parent(color, ans[color][x]);
}
void gabung(int color, int x, int y) {
  int a = parent(color, x), b = parent(color, y);
  if (a != b) ans[color][a] = b;
}
int main() {
  int n, m, q, a, b, c, total;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) ans[i][j] = j;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &a, &b, &c);
    gabung(c, a, b);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    total = 0;
    scanf("%d %d", &a, &b);
    for (int j = 1; j <= m; j++)
      total += (parent(j, a) == parent(j, b) ? 1 : 0);
    printf("%d\n", total);
  }
  return 0;
}
