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;
vector<vector<pair<int, int> > > graph(101);
bool dfs(int u, int v, int c) {
  int x, y;
  bool visited[101];
  for (int i = 0; i < 101; i++) visited[i] = false;
  stack<int> st;
  st.push(u);
  visited[u] = true;
  while (!st.empty()) {
    x = st.top();
    st.pop();
    vector<pair<int, int> > adjLst = graph[x];
    for (int i = 0; i < adjLst.size(); i++) {
      y = adjLst[i].first;
      if (adjLst[i].second == c && !visited[y]) {
        visited[y] = true;
        st.push(y);
        if (y == v) return true;
      }
    }
  }
  return false;
}
int main() {
  int n, m, q, a, b, c, cnt;
  ;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    graph[a].push_back({b, c});
    graph[b].push_back({a, c});
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    cnt = 0;
    cin >> a >> b;
    for (int k = 1; k <= m; k++) {
      if (dfs(a, b, k)) cnt++;
    }
    cout << cnt << endl;
    ;
  }
}
