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;
long long mod = 1;
int n, m, x, y, u, v, q;
vector<pair<int, int> > ve[105];
int par[105];
int vs[105];
int cnt;
map<pair<int, int>, int> mp;
bool f[105][105][105];
int test;
bool ok;
int solu(int node) {
  int i, j;
  vs[node] = cnt;
  if (node == v) {
    ok = 1;
    return 1;
  }
  for (i = 0; i < ve[node].size(); i++) {
    int rr = ve[node][i].first;
    int now = ve[node][i].second;
    if (vs[rr] != cnt && f[node][rr][test]) {
      solu(rr);
    }
  }
  return 0;
}
int main() {
  int i, j, t, tc = 1, a, b, c;
  scanf("%d %d", &n, &m);
  for (i = 0; i < m; i++) {
    scanf("%d %d %d", &u, &v, &c);
    ve[u].push_back(pair<int, int>(v, c));
    ve[v].push_back(pair<int, int>(u, c));
    f[u][v][c] = 1;
    f[v][u][c] = 1;
  }
  for (i = 1; i <= 105; i++) par[i] = i;
  scanf("%d", &q);
  int ans = 0;
  while (q--) {
    scanf("%d %d", &u, &v);
    for (i = 1; i <= 105; i++) {
      cnt = i;
      test = i;
      ok = 0;
      int res = solu(u);
      if (ok) ans++;
    }
    cout << ans << endl;
    ok = 0;
    ans = 0;
    memset(vs, 0, sizeof vs);
  }
  return 0;
}
