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;
unordered_map<int, int> H[100007], ans[100007];
int fi(int x, int y) { return x == H[x][y] ? x : H[x][y] = fi(H[x][y], y); }
void uni(int x, int y, int c) {
  if (H[x].find(c) == H[x].end()) H[x][c] = x;
  if (H[y].find(c) == H[y].end()) H[y][c] = y;
  int p = fi(x, c), q = fi(y, c);
  if (p == q) return;
  if (p > q) swap(p, q);
  H[fi(x, c)][c] = fi(y, c);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (x > y) swap(x, y);
    uni(x, y, z);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (H[x].size() > H[y].size()) swap(x, y);
    if (ans[x].find(y) == ans[x].end()) {
      int Ans = 0;
      for (auto &c : H[x]) {
        if (H[y].find(c.first) == H[y].end()) continue;
        int p = fi(x, c.first), q = fi(y, c.first);
        if (p == q) Ans++;
      }
      ans[x][y] = Ans;
    }
    printf("%d\n", ans[x][y]);
  }
}
