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 = 4 * 1000 * 1000 + 100;
vector<int> vec[maxn];
pair<pair<int, int>, int> edge[maxn];
int pre[maxn], par[maxn], n, m;
map<pair<int, int>, int> ans;
int get_par(int v) { return par[v] = (par[v] == v ? v : get_par(par[v])); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 1000000; i++) par[i] = i;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int f, g, c;
    cin >> f >> g >> c;
    vec[f].push_back(c);
    vec[g].push_back(c);
    edge[i] = {{f, g}, c};
  }
  for (int i = 1; i <= n; i++) {
    sort(vec[i].begin(), vec[i].end());
    vec[i].erase(unique(vec[i].begin(), vec[i].end()), vec[i].end());
    pre[i] = pre[i - 1] + ((int)vec[i - 1].size());
  }
  for (int i = 0; i < m; i++) {
    int fr = edge[i].first.first;
    int sec = edge[i].first.second;
    int c = edge[i].second;
    int ind1 = pre[fr] + lower_bound(vec[fr].begin(), vec[fr].end(), c) -
               vec[fr].begin();
    int ind2 = pre[sec] + lower_bound(vec[sec].begin(), vec[sec].end(), c) -
               vec[sec].begin();
    par[get_par(ind1)] = get_par(ind2);
  }
  int q;
  cin >> q;
  while (q--) {
    int f, g;
    cin >> f >> g;
    if (((int)vec[f].size()) > ((int)vec[g].size())) swap(f, g);
    if (ans.count({f, g})) {
      cerr << "                                  ";
      cout << ans[make_pair(f, g)] << endl;
    } else {
      int anse = 0;
      for (int i = 0; i < ((int)vec[f].size()); i++) {
        int c = vec[f][i];
        if (lower_bound(vec[g].begin(), vec[g].end(), c) != vec[g].end()) {
          assert(pre[f] >= 0 && pre[g] >= 0);
          int ind1 = pre[f] + i;
          int ind2 = pre[g] + (lower_bound(vec[g].begin(), vec[g].end(), c) -
                               vec[g].begin());
          if (get_par(ind1) == get_par(ind2)) anse++;
        }
      }
      ans[make_pair(f, g)] = anse;
      cerr << "                    ";
      cout << anse << endl;
    }
  }
}
