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 = 100005;
unordered_map<int, int> par[MAXN], mp[MAXN], size[MAXN];
int getpar(int x, int c) {
  if (par[x][c] == x) return x;
  return par[x][c] = getpar(par[x][c], c);
}
inline void add(int x, int y, int c) {
  if (par[x].count(c) == 0) par[x][c] = x;
  if (par[y].count(c) == 0) par[y][c] = y;
  x = getpar(x, c), y = getpar(y, c);
  if (x == y) return;
  if (size[x][c] > size[y][c]) swap(x, y);
  size[y][c] += size[x][c] + 1;
  par[x][c] = y;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    add(a, b, c);
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (size[x].size() > par[y].size()) swap(x, y);
    if (mp[x].count(y)) {
      cout << mp[x][y] << ' ';
      continue;
    }
    int ans = 0;
    for (auto &a : par[x])
      if (par[y].count(a.first) != 0 &&
          getpar(x, a.first) == getpar(y, a.first))
        ans++;
    mp[x][y] = ans;
    cout << ans << ' ';
  }
  cout << endl;
  return 0;
}
