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;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
vector<pair<long long, long long>> adj[101];
vector<long long> vis(101, 0);
bool dfs(long long u, long long c, long long dest) {
  vis[u] = 1;
  if (u == dest) return true;
  for (auto v : adj[u]) {
    if (!vis[v.second] && v.first == c) {
      if (dfs(v.second, c, dest)) return true;
    }
  }
  return false;
}
void solve() {
  long long n, m, q;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long x, y, c;
    cin >> x >> y >> c;
    adj[x].push_back({c, y});
    adj[y].push_back({c, x});
  }
  cin >> q;
  while (q--) {
    long long a, b, cnt = 0;
    cin >> a >> b;
    for (long long i = 1; i < m + 1; i++) {
      vis.assign(101, 0);
      if (dfs(a, i, b)) cnt++;
    }
    cout << cnt << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  for (long long i = 1; i < t + 1; i++) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
