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 OO = (int)1e5;
const double PI = acos(-1.0);
const double EPS = (1e-9);
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<pair<int, int> > g[110];
int n, m, a, b, c, q, u, v;
bool vis[110];
int path(int x, int y, int clr) {
  vis[x] = 1;
  queue<int> q;
  q.push(x);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    if (cur == y) return 1;
    for (auto i : g[cur])
      if (i.second == clr && !vis[i.first]) q.push(i.first), vis[i.first] = 1;
  }
  return 0;
}
int main() {
  fast();
  cin >> n >> m;
  for (int(i) = 0; (i) < (int)m; (i)++) {
    cin >> a >> b >> c;
    g[a].push_back({b, c});
    g[b].push_back({a, c});
  }
  cin >> q;
  while (q--) {
    cin >> u >> v;
    int ans = 0;
    for (int i = 1; i <= 100; i++) {
      memset(vis, 0, sizeof(vis));
      ;
      ans += path(u, v, i);
    }
    cout << ans << endl;
  }
}
