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<bool> visited;
vector<bool> visitedColor;
vector<vector<pair<int, int>>> V;
int ans;
void dfs(int curr, int v, int color) {
  if (curr == v) {
    if (!visitedColor.at(color)) {
      visitedColor.at(color) = true;
      ++ans;
    }
    return;
  }
  visited.at(curr) = true;
  for (auto [node, c] : V.at(curr)) {
    if ((c == color) && !visited.at(node)) {
      dfs(node, v, color);
    }
  }
  visited.at(curr) = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  V = vector<vector<pair<int, int>>>(n + 1);
  visited = vector<bool>(n + 1);
  int n1, n2, c;
  for (size_t i = 0; i < m; i++) {
    cin >> n1 >> n2 >> c;
    V.at(n1).push_back(make_pair(n2, c));
    V.at(n2).push_back(make_pair(n1, c));
  }
  int q, u, v;
  cin >> q;
  for (size_t i = 0; i < q; i++) {
    visitedColor = vector<bool>(m + 1);
    cin >> u >> v;
    ans = 0;
    visited.at(u) = true;
    for (auto [node, color] : V.at(u)) {
      dfs(node, v, color);
    }
    visited.at(u) = false;
    cout << ans << endl;
  }
  return 0;
}
