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 long long N = 100005;
map<long long, long long> colors[N];
map<pair<long long, long long>, long long> done;
long long par[N];
inline long long get(long long x) {
  return ((par[x] == x) ? (x) : (par[x] = get(par[x])));
}
inline void unite(long long v, long long u) {
  long long parv = get(v);
  long long paru = get(u);
  par[paru] = parv;
  return;
}
inline bool same(long long v, long long u) {
  long long parv = get(v);
  long long paru = get(u);
  return (parv == paru);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m, v, u, i, c, q, ans, a, b;
  cin >> n >> m;
  for (i = 1; i < N; i++) par[i] = i;
  for (i = 1; i <= m; i++) {
    cin >> v >> u >> c;
    a = b = -1;
    if (colors[v].count(c)) a = colors[v][c];
    if (colors[u].count(c)) b = colors[u][c];
    if (a != -1) unite(a, i);
    if (b != -1) unite(b, i);
    colors[v][c] = i;
    colors[u][c] = i;
  }
  cin >> q;
  while (q--) {
    cin >> v >> u;
    if (done.count(make_pair(v, u)))
      cout << done[make_pair(v, u)] << "\n";
    else {
      ans = 0;
      if (colors[v].size() > colors[u].size()) swap(v, u);
      for (map<long long, long long>::iterator vit = colors[v].begin();
           vit != colors[v].end(); vit++) {
        map<long long, long long>::iterator uit = colors[u].find(vit->first);
        if (uit != colors[v].end()) ans += (same(vit->second, uit->second));
      }
      done[make_pair(v, u)] = done[make_pair(u, v)] = ans;
      cout << ans << "\n";
    }
  }
  return 0;
}
