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 = 1000 * 100 + 10;
unordered_map<int, int> p[MAXN], ans[MAXN];
int get_par(int v, int c) {
  if (p[v].find(c) == p[v].end() || p[v][c] == v) return p[v][c] = v;
  return p[v][c] = get_par(p[v][c], c);
}
void merge(int u, int v, int c) {
  u = get_par(u, c), v = get_par(v, c);
  if (u != v) p[u][c] = v;
  return;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    merge(a - 1, b - 1, c);
  }
  int q;
  cin >> q;
  for (int i = 0; i < n; i++)
    for (auto it : p[i]) p[i][it.first] = get_par(it.second, it.first);
  while (q--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    if ((int)p[a].size() > (int)p[b].size())
      swap(a, b);
    else if ((int)p[a].size() == (int)p[b].size() && a > b)
      swap(a, b);
    auto x = ans[a].find(b);
    if (x != ans[a].end()) {
      cout << x->second << endl;
      continue;
    }
    int anss = 0;
    for (auto it : p[a]) {
      auto itt = p[b].find(it.first);
      if (itt != p[b].end() && it.second == itt->second) anss++;
    }
    ans[a][b] = anss;
    cout << anss << endl;
  }
  return 0;
}
