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;
unordered_map<int, int> P[100100];
set<int> C[100100];
vector<int> L[100100];
map<pair<int, int>, int> ans;
int find(int c, int x) {
  if (P[c].find(x) == P[c].end()) {
    P[c][x] = x;
    return x;
  }
  if (P[c][x] == x) return x;
  P[c][x] = find(c, P[c][x]);
  return P[c][x];
}
int main() {
  int N, M, Q, i, j, k, n1, n2, n3, x, y, z;
  scanf("%d%d", &N, &M);
  for (i = 0; i < M; i++) {
    cin >> n1 >> n2 >> n3;
    n1--;
    n2--;
    n3--;
    if (n1 > n2) swap(n1, n2);
    P[n3][find(n3, n1)] = find(n3, n2);
    C[n1].insert(n3);
    C[n2].insert(n3);
  }
  cin >> Q;
  set<int>::iterator it;
  for (i = 0; i < N; i++) {
    for (it = C[i].begin(); it != C[i].end(); ++it) {
      L[i].push_back((*it));
    }
  }
  for (i = 0; i < Q; i++) {
    scanf("%d%d", &n1, &n2);
    n1--;
    n2--;
    if (L[n1].size() > L[n2].size()) swap(n1, n2);
    if (ans.find(pair<int, int>(n1, n2)) != ans.end()) {
      cout << ans[pair<int, int>(n1, n2)] << endl;
    } else {
      int res = 0;
      for (j = 0; j < L[n1].size(); j++) {
        k = L[n1][j];
        if (P[k].find(n2) == P[k].end()) continue;
        if (find(k, n1) == find(k, n2)) res++;
      }
      ans[pair<int, int>(n1, n2)] = res;
      cout << res << endl;
    }
  }
  return 0;
}
