Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5

#include <bits/stdc++.h>
using namespace std;
int X[] = {1, -1, 0, 0};
int Y[] = {0, 0, 1, -1};
const int N = 2e5 + 5;
int n, m, q;
vector<vector<int> > graph(101);
map<pair<int, int>, vector<int> > clr;
vector<vector<int> > ans(101, vector<int>(101, 0));
void bfs(int pos) {
  queue<int> q;
  q.push(pos);
  vector<vector<int> > color(n + 1, vector<int>(m + 1));
  for (int i = 1; i <= m; i++) color[pos][i] = 1;
  vector<bool> vis(n + 1);
  vis[pos] = 1;
  int now = 101;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int to : graph[u]) {
      if (!vis[to]) {
        vis[to] = 1;
        q.push(to);
        if (now == 0) continue;
        int cnt = 0;
        vector<int> tmp = clr[make_pair(min(to, u), max(to, u))];
        for (auto it : tmp) (color[u][it] == 1) ? color[to][it] = 1 : 1;
        for (int i = 1; i <= m; i++) {
          if (color[to][i]) cnt++;
        }
        ans[to][pos] = ans[pos][to] = cnt;
        now = cnt;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  int a, b, c;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    graph[a].push_back(b);
    graph[b].push_back(a);
    if (a > b) swap(a, b);
    clr[make_pair(a, b)].push_back(c);
  }
  for (int i = 1; i <= n; i++) bfs(i);
  cin >> q;
  while (q--) {
    cin >> a >> b;
    cout << ans[a][b] << endl;
  }
}
