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;
void input() {}
vector<int> v[120];
vector<int> col[120][120];
bool vi[120];
bool ch[120][120];
bool cchk[120];
int a, b, ans;
void dfs(int i, int c) {
  vi[i] = 1;
  if (i == b) {
    if (!cchk[c]) {
      cchk[c] = 1;
      ans++;
    }
    vi[i] = 0;
    return;
  }
  for (int j = 0; j < v[i].size(); ++j) {
    if (!vi[v[i][j]]) {
      int k;
      for (k = 0; k < col[i][v[i][j]].size(); ++k) {
        if (col[i][v[i][j]][k] == c) dfs(v[i][j], c);
        ;
      }
    }
  }
  vi[i] = 0;
  return;
}
int main() {
  input();
  for (int i = 0; i < 1000000; ++i) {
    for (int j = 0; j < 100000; ++j) {
    }
  }
  int n, m;
  cin >> n >> m;
  memset(ch, 0, sizeof(ch));
  for (int i = 0; i < m; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    if (!ch[x][y]) {
      ch[x][y] = 1;
      ch[y][x] = 1;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    col[x][y].push_back(c);
    col[y][x].push_back(c);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    memset(vi, 0, sizeof(vi));
    memset(cchk, 0, sizeof(cchk));
    cin >> a >> b;
    ans = 0;
    for (int j = 0; j < v[a].size(); ++j) {
      for (int k = 0; k < col[a][v[a][j]].size(); ++k) {
        vi[a] = 1;
        dfs(v[a][j], col[a][v[a][j]][k]);
        vi[a] = 0;
      }
    }
    cout << ans << endl;
  }
}
