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 MAXN = 1000 + 10;
vector<int> adj[MAXN][MAXN];
bool mark[MAXN];
long long int mod = 1000 * 1000 * 1000 + 7;
void addEdge(int x, int y, int c) {
  adj[c][x].push_back(y);
  adj[c][y].push_back(x);
}
void dfs(int u, int c) {
  mark[u] = true;
  for (int i = 0; i < adj[c][u].size(); i++) {
    int x = adj[c][u][i];
    if (!mark[x]) dfs(x, c);
  }
  return;
}
void cl(int n) {
  for (int i = 0; i <= n; i++) mark[i] = false;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    addEdge(x, y, c);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    int ans = 0;
    for (int j = 1; j <= m; j++) {
      cl(n);
      dfs(x, j);
      if (mark[y]) {
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
