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 int mx = 1e5 + 10;
vector<int> wt[mx], w[mx];
pair<int, int> mp[mx + mx];
int mpc = 0;
int us[mx], vs[mx], cs[mx];
int p[mx + mx], sz[mx + mx];
map<pair<int, int>, int> mem;
int gp(int u) {
  while (u != p[u]) u = p[u];
  return u;
}
void join(int u, int v) {
  int pu = gp(u);
  int pv = gp(v);
  if (pu == pv) return;
  if (sz[pu] < sz[pv]) swap(pu, pv);
  p[pv] = pu;
  sz[pu] += sz[pv];
}
inline int id(int u, int c) {
  pair<int, int> pu(u, c);
  int idx = lower_bound(mp, mp + mpc, pu) - mp;
  if (idx >= mpc) return -1;
  if (mp[idx] != pu) return -1;
  return idx;
}
int main() {
  int n, m, u, v, c, q;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &u, &v, &c);
    u--, v--;
    wt[u].push_back(c);
    wt[v].push_back(c);
    us[i] = u, vs[i] = v, cs[i] = c;
  }
  for (int i = 0; i < n; i++) {
    sort(wt[i].begin(), wt[i].end());
    int deg = wt[i].size();
    for (int j = 0; j < deg;) {
      c = wt[i][j];
      while (j < deg && wt[i][j] == c) j++;
      w[i].push_back(c);
      mp[mpc++] = make_pair(i, c);
    }
  }
  for (int i = 0; i < mx + mx; i++) p[i] = i, sz[i] = 1;
  for (int i = 0; i < m; i++) join(id(us[i], cs[i]), id(vs[i], cs[i]));
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &u, &v);
    u--, v--;
    if (w[u].size() > w[v].size()) swap(u, v);
    int ans = 0;
    pair<int, int> qry(u, v);
    if (mem.count(qry) == 1)
      ans = mem[qry];
    else {
      for (int i = 0; i < int(w[u].size()); i++) {
        int c = w[u][i];
        int idu = id(u, c);
        int idv = id(v, c);
        if (idv != -1 && gp(idu) == gp(idv)) ans++;
      }
      mem[qry] = ans;
    }
    printf("%d\n", ans);
  }
  return 0;
}
