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;
map<int, int> NCE[100009];
int n, m, a, b, c, grp[100009];
int GetGroup(int a) { return grp[a] == a ? a : (grp[a] = GetGroup(grp[a])); }
void Combine(int a, int b) {
  grp[GetGroup(a)] = GetGroup(b);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < m; i++) grp[i] = i;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    int e1 = -1, e2 = -1;
    if (NCE[a].count(c)) e1 = NCE[a][c];
    if (NCE[b].count(c)) e2 = NCE[b][c];
    if (e1 >= 0) Combine(e1, i);
    if (e2 >= 0) Combine(i, e2);
    NCE[a][c] = i;
    NCE[b][c] = i;
  }
  int q;
  cin >> q;
  map<pair<int, int>, int> cache;
  while (q--) {
    cin >> a >> b;
    if (cache.count({a, b}) != 0) {
      cout << cache[{a, b}] << endl;
      continue;
    }
    int sa = NCE[a].size(), sb = NCE[b].size();
    if (sa > sb) swap(sa, sb), swap(a, b);
    auto &B = NCE[b];
    int ans = 0;
    for (auto it = NCE[a].begin(); it != NCE[a].end(); it++) {
      auto A = B.find(it->first);
      if (A == B.end()) continue;
      if (GetGroup(A->second) == GetGroup(it->second)) ans++;
    }
    cout << ans << endl;
    cache[{a, b}] = cache[{b, a}] = ans;
  }
  return 0;
}
