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;
using ll = long long;
using db = double;
using ld = long double;
using str = string;
template <class T>
using V = vector<T>;
template <class T1, class T2>
using umap = unordered_map<T1, T2>;
using pi = pair<int, int>;
using vi = V<int>;
int n, e, src, dest;
V<bool> visited;
V<V<pair<int, int>>> adj;
bool dfs(int node, int color);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> e;
  adj.resize(n);
  for (int i = 0; i < e; i++) {
    int a, b, col;
    cin >> a >> b >> col;
    a--;
    b--;
    col--;
    adj[a].push_back({b, col});
    adj[b].push_back({a, col});
  }
  int q;
  cin >> q;
  while (q--) {
    cin >> src >> dest;
    src--;
    dest--;
    int cnt = 0;
    for (int i = 0; i < e; i++) {
      visited.clear();
      visited.resize(n, 0);
      cnt += dfs(src, i);
    }
    cout << cnt << "\n";
  }
}
bool dfs(int node, int color) {
  visited[node] = 1;
  if (node == dest) return 1;
  for (auto u : adj[node])
    if (u.second == color && !visited[u.first])
      if (dfs(u.first, u.second)) return true;
  return false;
}
