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<pair<int, int>> count_freq(vector<int>& v) {
  int count = 1;
  vector<pair<int, int>> res;
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size(); i++) {
    if (v[i] != v[i - 1]) {
      res.push_back(make_pair(v[i - 1], count));
      count = 1;
    } else {
      count++;
    }
  }
  res.push_back(make_pair(v[v.size() - 1], count));
  return res;
}
vector<int> vector_input(int n) {
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  return v;
}
bool bfs(vector<vector<int>>& g, int u, int v, int n) {
  vector<int> visited(n + 1);
  queue<int> q;
  q.push(u);
  visited[u] = 1;
  while (!q.empty()) {
    int f = q.front();
    q.pop();
    for (auto i = g[f].begin(); i != g[f].end(); i++) {
      if (!visited[*i]) {
        q.push(*i);
        visited[*i] = 1;
      }
    }
  }
  return visited[v] == 1 ? 1 : 0;
}
void solve() {
  int n, m;
  cin >> n >> m;
  int d1 = m + 1;
  int d2 = n + 1;
  vector<vector<vector<int>>> v(m + 1);
  for (int i = 0; i < v.size(); i++) {
    vector<vector<int>> temp(n + 1);
    v[i] = temp;
  }
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    v[c][a].push_back(b);
    v[c][b].push_back(a);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    int count = 0;
    for (int j = 0; j < v.size(); j++) {
      if (v[j].size() == 0) continue;
      auto res = bfs(v[j], x, y, n);
      if (res) count++;
    }
    cout << count << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}
