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;
const int inf = numeric_limits<int>::max();
const long long linf = numeric_limits<long long>::max();
const int max_n = 101;
int n, m, q;
bool g[max_n][max_n][max_n], visited[max_n];
bool dfs(int st, int end, int col, int par = -1) {
  if (st == end) return true;
  if (visited[st]) return false;
  visited[st] = true;
  bool ret = false;
  for (int i = 0; i < (max_n); ++i) {
    if (g[st][i][col] && i != par) ret |= dfs(i, end, col, st);
  }
  return ret;
}
int main() {
  int u, v, c;
  cin >> n;
  cin >> m;
  for (int i = 0; i < (m); ++i) {
    cin >> u, cin >> v, cin >> c;
    g[u][v][c] = true;
    g[v][u][c] = true;
  }
  cin >> q;
  for (int i = 0; i < (q); ++i) {
    cin >> u, cin >> v;
    c = 0;
    for (int j = (1); j < (m); ++j) {
      fill(visited, visited + max_n, false);
      c += dfs(u, v, j);
    }
    cout << c << endl;
  }
  return 0;
}
