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 inf = 1e9 + 5;
const int nax = 1e5 + 5;
int n, m, a, b, c, q;
int ojc[nax], e[nax], f[nax], stan[nax];
vector<pair<int, int> > v[nax], dupa[nax];
map<pair<int, int>, int> mapa;
int find(int f) {
  if (ojc[f] == f) return f;
  ojc[f] = find(ojc[f]);
  return ojc[f];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i <= ((m)-1); ++i) {
    scanf("%d%d%d", &a, &b, &c);
    v[c].push_back(make_pair(a, b));
  }
  scanf("%d", &q);
  for (int i = (0); i <= ((q)-1); ++i) scanf("%d%d", &e[i], &f[i]);
  for (int i = (1); i <= (n); ++i) ojc[i] = i;
  for (int i = (1); i <= (m); ++i) {
    for (auto j : v[i]) {
      a = find(j.first);
      b = find(j.second);
      ojc[a] = b;
    }
    for (auto j : v[i]) {
      a = find(j.first);
      b = find(j.second);
      if (stan[j.first] != i) {
        stan[j.first] = i;
        dupa[j.first].push_back(make_pair(i, a));
      }
      if (stan[j.second] != i) {
        stan[j.second] = i;
        dupa[j.second].push_back(make_pair(i, b));
      }
    }
    for (auto j : v[i]) {
      ojc[j.first] = j.first;
      ojc[j.second] = j.second;
    }
  }
  for (int i = (0); i <= ((q)-1); ++i) {
    if (dupa[e[i]].size() > dupa[f[i]].size()) swap(e[i], f[i]);
    if (mapa.find(make_pair(e[i], f[i])) == mapa.end()) {
      int res = 0;
      auto x = dupa[f[i]].begin();
      for (auto j : dupa[e[i]]) {
        x = lower_bound(x, dupa[f[i]].end(), j);
        if (x == dupa[f[i]].end()) break;
        res += *x == j;
      }
      printf("%d\n", res);
      mapa[make_pair(e[i], f[i])] = res;
      mapa[make_pair(f[i], e[i])] = res;
    } else
      printf("%d\n", mapa[make_pair(e[i], f[i])]);
  }
  return 0;
}
