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;
#pragma GCC target("sse4")
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  int n, m;
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  static vector<pair<int, int> > v[1 << 20];
  static int f[1 << 20];
  function<int(int)> get = [&](int k) {
    return f[k] == k ? k : f[k] = get(f[k]);
  };
  static unordered_map<int, int> P[1 << 20];
  static int F[1 << 20];
  static int ans[1 << 20];
  static pair<int, int> Query[1 << 20];
  while (m--) {
    int a, b, c;
    cin >> a >> b >> c;
    v[c].push_back({a, b});
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int a, b;
    cin >> a >> b;
    Query[i] = {a, b};
    if (P[a].count(b))
      F[i] = P[a][b];
    else
      P[a][b] = P[b][a] = i;
  }
  static int when[1 << 20];
  for (int cnt = 1; cnt <= 1e5; ++cnt) {
    vector<int> Sort;
    for (auto it : v[cnt])
      when[it.first] = when[it.second] = cnt, f[it.first] = it.first,
      f[it.second] = it.second, Sort.push_back(it.first),
      Sort.push_back(it.second);
    sort(Sort.begin(), Sort.end());
    Sort.resize(unique(Sort.begin(), Sort.end()) - Sort.begin());
    for (auto it : v[cnt]) f[get(it.first)] = get(it.second);
    int sz = Sort.size();
    if (sz <= 400)
      for (int i = 0; i < sz; ++i)
        for (int j = i + 1; j < sz; ++j) {
          int a = Sort[i], b = Sort[j];
          if (get(a) == get(b) && P[a].count(b)) ans[P[a][b]]++;
        }
    else
      for (int i = 1; i <= q; ++i) {
        int a = Query[i].first, b = Query[i].second;
        if (get(a) == get(b) && when[a] == cnt && when[b] == cnt) ++ans[i];
      }
  }
  for (int i = 1; i <= q; ++i)
    if (F[i]) ans[i] = ans[F[i]];
  for (int i = 1; i <= q; ++i) cout << ans[i] << '\n';
  cout << '\n';
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}
