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 long long MOD = 1e9 + 7, MAXN = 1e5 + 10, INF = 1e18;
template <long long SIZE>
struct DSU {
  unordered_map<long long, long long> par[SIZE];
  unordered_map<long long, long long> sz[SIZE];
  long long Find(long long v, long long c) {
    return (par[v][c] <= 0 ? v : par[v][c] = Find(par[v][c], c));
  }
  long long Union(long long v, long long u, long long c) {
    long long pv = Find(v, c), pu = Find(u, c);
    if (pv == pu) return 0;
    if (par[pu][c] > par[pv][c]) swap(pu, pv);
    par[pv][c] += par[pu][c];
    par[pu][c] = pv;
    sz[pv][c] += sz[pu][c];
    return 1;
  }
};
DSU<MAXN> dsu;
unordered_map<long long, long long> mp[MAXN];
long long n, m, q, u, v, ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    dsu.Union(a, b, c--);
  }
  cin >> q;
  while (q--) {
    cin >> u >> v;
    if (dsu.par[u].size() > dsu.par[v].size()) swap(u, v);
    if (dsu.sz[u].size() > dsu.sz[v].size()) swap(u, v);
    if (mp[v].find(u) != mp[v].end()) mp[u][v] = mp[v][u];
    if (mp[u].find(v) == mp[u].end()) {
      long long ans = mp[u][v];
      for (auto t : dsu.par[u]) {
        if (dsu.par[v].find(t.first) != dsu.par[v].end()) {
          if (dsu.Find(u, t.first) == dsu.Find(v, t.first)) ans++;
        }
      }
      mp[u][v] = ans;
    }
    cout << mp[u][v] << '\n';
  }
  return 0;
}
