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;
vector<pair<int, int> > adj[101];
queue<pair<int, int> > qu;
map<pair<int, int>, int> ans;
set<pair<int, int> > chk[100];
int bfs(int en) {
  int x, cnt, i, y, co, pr, d;
  cnt = d = 0;
  while (!qu.empty()) {
    x = qu.front().first;
    pr = qu.front().second;
    qu.pop();
    for (i = 0; i < adj[x].size(); i++) {
      y = adj[x][i].first;
      co = adj[x][i].second;
      if ((chk[co].find({x, y}) != chk[co].end()) || (co != pr && pr != -1))
        continue;
      chk[co].insert({x, y});
      chk[co].insert({y, x});
      if (y == en) {
        cnt++;
      } else
        qu.push({y, co});
    }
  }
  return cnt;
}
int main() {
  int n, m, i, q, a, b, c;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    scanf("%d %d %d", &a, &b, &c);
    adj[a].push_back({b, c});
    adj[b].push_back({a, c});
  }
  cin >> q;
  while (q--) {
    scanf("%d %d", &a, &b);
    if (ans.find({a, b}) == ans.end()) {
      qu.push({a, -1});
      c = bfs(b);
      if (c == 0) c = -1;
      ans[{a, b}] = c;
      ans[{b, a}] = c;
      for (i = 0; i < 101; i++) chk[i].clear();
    } else {
      c = ans[{a, b}];
    }
    if (c == -1) c = 0;
    printf("%d\n", c);
  }
  return 0;
}
