Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
const long long N = 1004;
vector<pair<long long, long long>> gr[N];
long long vis[N] = {};
bool dfs(long long curr, long long &col, long long &dest, long long par) {
  if (curr == dest) return true;
  vis[curr] = 1;
  bool ans = 0;
  for (auto x : gr[curr]) {
    if (!vis[x.first] and x.first != par and x.second == col) {
      ans = ans or dfs(x.first, col, dest, curr);
    }
  }
  return ans;
}
void solve(long long tc) {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    gr[i].clear();
  }
  long long x, y, c;
  for (long long i = 0; i < m; i++) {
    cin >> x >> y >> c;
    gr[x].push_back({y, c});
    gr[y].push_back({x, c});
  }
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> x >> y;
    long long ans = 0;
    for (long long j = 0; j < gr[x].size(); j++) {
      memset(vis, 0, sizeof(vis));
      vis[x] = 1;
      ans += dfs(gr[x][j].first, gr[x][j].second, y, x);
    }
    cout << ans << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve(1);
  return 0;
}
