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 = 100 * 1000 + 10;
vector<pair<int, int> > v[maxn];
int par[maxn], ans[maxn], aray[maxn * 2];
vector<int> w[maxn];
pair<int, int> Q[maxn];
map<int, int> mp[maxn];
inline int find_par(int id) {
  if (par[id] < 0) return id;
  return par[id] = find_par(par[id]);
}
inline void add_edge(int fi, int se) {
  int u = find_par(fi), v = find_par(se);
  if (u == v) return;
  if (par[u] < par[v]) swap(u, v);
  par[v] += par[u];
  par[u] = v;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  memset(par, -1, sizeof par);
  for (int i = 1; i <= m; ++i) {
    int fi, se, col;
    cin >> fi >> se >> col;
    v[col].push_back({fi, se});
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) cin >> Q[i].first >> Q[i].second;
  for (int i = 1; i <= m; ++i) {
    if (v[i].size()) {
      for (int j = 0; j < v[i].size(); ++j)
        add_edge(v[i][j].first, v[i][j].second);
      if (v[i].size() >= 500) {
        for (int j = 1; j <= q; ++j)
          if (find_par(Q[j].first) == find_par(Q[j].second)) ans[j]++;
      } else {
        int cur = 0;
        for (int j = 0; j < v[i].size(); ++j) {
          aray[cur++] = v[i][j].first;
          aray[cur++] = v[i][j].second;
        }
        sort(aray, aray + cur);
        cur = unique(aray, aray + cur) - aray;
        for (int k = 0; k < cur; ++k)
          for (int l = k + 1; l < cur; ++l)
            if (find_par(aray[k]) == find_par(aray[l])) {
              mp[aray[k]][aray[l]]++;
              mp[aray[l]][aray[k]]++;
            }
      }
      for (int j = 0; j < v[i].size(); ++j)
        par[v[i][j].first] = par[v[i][j].second] = -1;
    }
  }
  for (int i = 1; i <= q; ++i)
    cout << ans[i] + mp[Q[i].first][Q[i].second] << endl;
  return 0;
}
