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;
vector<vector<pair<int, int> > > adj;
int n;
bool bfs(int src, int target, int color) {
  vector<int> dist(n + 1, 100000);
  queue<pair<int, int> > q;
  dist[src] = 0;
  q.push(make_pair(src, 0));
  while (!q.empty()) {
    int cur = q.front().first;
    int d = q.front().second;
    q.pop();
    for (int i = 0; i < adj[cur].size(); i++) {
      if (dist[adj[cur][i].first] == 100000 && adj[cur][i].second == color) {
        dist[adj[cur][i].first] = d + 1;
        q.push(make_pair(adj[cur][i].first, d + 1));
        if (adj[cur][i].first == target) return true;
      }
    }
  }
  return false;
}
void make_edge(int x1, int x2, int c) {
  adj[x1].push_back(make_pair(x2, c));
  adj[x2].push_back(make_pair(x1, c));
}
int main() {
  cin >> n;
  int m;
  cin >> m;
  adj.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int x1, x2, x3;
    cin >> x1 >> x2 >> x3;
    make_edge(x1, x2, x3);
  }
  int q;
  cin >> q;
  int cnt = 0;
  for (int i = 0; i < q; i++) {
    int x1, x2;
    cin >> x1 >> x2;
    for (int j = 1; j <= m; j++) {
      if (bfs(x1, x2, j)) cnt++;
    }
    cout << cnt << endl;
    cnt = 0;
  }
  return 0;
}
