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 INF = 100009;
unordered_map<int, int> f[INF], old[INF];
int n, m, q;
inline int find(int x, int c) {
  if (f[x][c] < 0) return x;
  return f[x][c] = find(f[x][c], c);
}
inline void merge(int x, int y, int c) {
  if (f[x].find(c) == f[x].end()) f[x][c] = -1;
  if (f[y].find(c) == f[y].end()) f[y][c] = -1;
  x = find(x, c), y = find(y, c);
  if (x == y) return;
  f[y][c] += f[x][c];
  f[x][c] = y;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1, x, y, c; i <= m; i++) {
    scanf("%d %d %d", &x, &y, &c);
    merge(x, y, c);
  }
  scanf("%d", &q);
  int a, b;
  while (q--) {
    scanf("%d %d", &a, &b);
    if (f[a].size() > f[b].size()) swap(a, b);
    if (old[a].find(b) == old[a].end()) {
      int ans = 0;
      for (auto &c : f[a])
        if (f[b].find(c.first) != f[b].end() &&
            find(a, c.first) == find(b, c.first))
          ans++;
      old[a][b] = ans;
    }
    printf("%d\n", old[a][b]);
  }
}
