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 dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int di[] = {1, -1, 0, 0};
int dj[] = {0, 0, 1, -1};
void file() {}
int n, m, u, v, c, vis[101], q;
vector<pair<int, int> > arr[101];
int col;
int ans[101];
void dfs(int x, int y, int clr) {
  if (vis[x]) return;
  vis[x] = 1;
  if (x == y) {
    ans[clr] = 1;
    return;
  }
  for (int i = 0; i < ((int)((arr[x]).size())); i++) {
    if (arr[x][i].second == clr) dfs(arr[x][i].first, y, clr);
  }
}
int main() {
  file();
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &u, &v, &c);
    arr[u].push_back({v, c});
    arr[v].push_back({u, c});
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &u, &v);
    for (int i = 0; i < ((int)((arr[u]).size())); i++) {
      dfs(u, v, arr[u][i].second);
      memset(vis, 0, sizeof vis);
    }
    int cnt = 0;
    for (int i = 0; i <= m; i++) {
      cnt += ans[i];
    }
    printf("%d\n", cnt);
    memset(ans, 0, sizeof ans);
  }
}
