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;
const int M = 1e9 + 7;
const int N = 1e4;
long long po(int, int);
vector<pair<int, int>> e[N];
vector<bool> vis;
set<int> s;
void dfs(int u, int c, int tr) {
  vis[u] = 1;
  if (u == tr) {
    s.insert(c);
    return;
  }
  for (auto i : e[u]) {
    if (!vis[i.first] && i.second == c) {
      dfs(i.first, c, tr);
    }
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, colour;
    cin >> x >> y >> colour;
    e[x].push_back({y, colour});
    e[y].push_back({x, colour});
  }
  int q;
  cin >> q;
  while (q--) {
    int start, target;
    cin >> start >> target;
    s.clear();
    for (auto i : e[start]) {
      vis.assign(n + 1, 0);
      vis[start] = 1;
      dfs(i.first, i.second, target);
    }
    cout << s.size() << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int i = 1; i < t + 1; i++) {
    solve();
  }
  return 0;
}
long long po(int a, int b) {
  if (b == 0) return 1;
  long long ans = 1;
  if (b % 2 == 0) {
    ans = po(a, b / 2) % M;
    ans = (ans % M * ans % M) % M;
  } else {
    ans = po(a, (b - 1) / 2) % M;
    ans = (ans % M * ans % M * a) % M;
  }
  return ans % M;
}
