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;
struct edge {
  int x, y, c;
} ee[110];
int n, m;
int cnt = 0;
int pa[110];
int vis[110];
int find(int x) {
  if (pa[x] != x) pa[x] = find(pa[x]);
  return pa[x];
}
void sert() {
  for (int i = 0; i <= n; i++) pa[i] = i;
}
bool kru(int c, int s, int e) {
  for (int i = 0; i < m; i++) {
    if (ee[i].c == c) {
      int x = find(ee[i].x);
      int y = find(ee[i].y);
      if (x != y) {
        pa[x] = y;
      }
      if (find(s) == find(e)) return true;
    }
  }
  return false;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    memset(vis, 0, sizeof(vis));
    int t;
    for (int i = 0; i < m; i++) {
      int x, y, c;
      scanf("%d%d%d", &x, &y, &c);
      vis[c] = 1;
      ee[i].x = x;
      ee[i].y = y;
      ee[i].c = c;
    }
    scanf("%d", &t);
    sert();
    for (int i = 0; i < t; i++) {
      cnt = 0;
      int st, ed;
      scanf("%d%d", &st, &ed);
      for (int i = 1; i <= m; i++) {
        if (vis[i]) {
          if (kru(i, st, ed)) cnt++;
        }
        sert();
      }
      printf("%d\n", cnt);
    }
  }
  return 0;
}
