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 maxn = 1e5 + 10, maxsq = 320 * 320;
vector<int> vert[maxn];
vector<pair<int, int> > edge[maxn], quer[maxn];
int par[maxn], ans[maxn];
int n, m, q;
int root(int v) { return (par[v] < 0) ? v : par[v] = root(par[v]); }
void merge(int v, int u) {
  if ((v = root(v)) == (u = root(u))) return;
  if (par[v] > par[u]) swap(v, u);
  par[v] += par[u];
  par[u] = v;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int fi, se, th;
    cin >> fi >> se >> th;
    th--;
    edge[th].push_back({fi, se});
    vert[th].push_back(fi);
    vert[th].push_back(se);
  }
  for (int i = 0; i < maxn; i++) {
    sort(vert[i].begin(), vert[i].end());
    vert[i].resize(unique(vert[i].begin(), vert[i].end()) - vert[i].begin());
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int fi, se;
    cin >> fi >> se;
    if (fi > se) swap(fi, se);
    quer[fi].push_back({se, i});
  }
  for (int i = 1; i <= n; i++) sort(quer[i].begin(), quer[i].end());
  for (int col = 0; col < maxn; col++) {
    for (auto v : vert[col]) par[v] = -1;
    for (auto e : edge[col]) merge(e.first, e.second);
    if (((int)(vert[col]).size()) >= maxsq) {
      for (auto v : vert[col]) {
        for (auto gar : quer[v]) {
          int u = gar.first;
          int id = lower_bound(vert[col].begin(), vert[col].end(), u) -
                   vert[col].begin();
          if (id != ((int)(vert[col]).size()) and vert[col][id] == u)
            ans[gar.second]++;
        }
      }
    } else {
      for (int i = 0; i < ((int)(vert[col]).size()); i++) {
        int v = vert[col][i];
        for (int j = i + 1; j < ((int)(vert[col]).size()); j++) {
          int u = vert[col][j];
          if (root(v) == root(u)) {
            int id =
                lower_bound(quer[v].begin(), quer[v].end(), make_pair(u, -1)) -
                quer[v].begin();
            while (id < ((int)(quer[v]).size()) and quer[v][id].first == u) {
              ans[quer[v][id].second]++;
              id++;
            }
          }
        }
      }
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
  return 0;
}
