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 double EXP = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x7fffffff;
const int MS = 105;
int fa[MS][MS];
void init() {
  for (int i = 0; i < MS; i++)
    for (int j = 0; j < MS; j++) fa[i][j] = j;
}
int find(int c, int a) {
  if (fa[c][a] == a) return a;
  return fa[c][a] = find(c, fa[c][a]);
}
void make_union(int c, int a, int b) {
  a = find(c, a);
  b = find(c, b);
  if (a == b) return;
  fa[c][b] = fa[c][a];
}
int main() {
  int n, m, q, a, b, c;
  init();
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    make_union(c, a, b);
  }
  cin >> q;
  while (q--) {
    int ans = 0;
    cin >> a >> b;
    for (c = 1; c <= m; c++)
      if (find(c, a) == find(c, b)) ans++;
    cout << ans << endl;
  }
  return 0;
}
