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;
int n, v, k, sum, m, ns, a, b, c;
vector<int> state(102);
struct edge {
  int a, b, c;
  edge() {}
  edge(int q, int w, int e) { a = q, b = w, c = e; }
};
vector<vector<edge> > gr;
vector<set<int> > col;
void dfs(int u, int color) {
  state[u] = 1;
  int k = gr[u].size();
  if (color != -1) col[u].insert(color);
  for (int i = 0; i < k; i++) {
    edge e = gr[u][i];
    if ((color == -1 || color == e.c) &&
        (state[e.b] == 0 || (state[e.b] == 1 && e.b == b && a == e.a))) {
      dfs(e.b, e.c);
      state[e.b] = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  gr.resize(n + 1);
  col.resize(n + 1);
  for (int i = 1; i <= m; i++) {
    cin >> a >> b >> c;
    gr[a].push_back(edge(a, b, c));
    gr[b].push_back(edge(b, a, c));
  }
  cin >> k;
  for (int i = 1; i <= k; i++) {
    cin >> a >> b;
    dfs(a, -1);
    cout << col[b].size() << '\n';
    state.assign(state.size(), 0);
    col.clear();
    col.resize(n + 1);
  }
  return 0;
}
