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 N = 1e5 + 10, sq = 455, msq = 240;
map<int, vector<int> > gr[N];
map<int, bool> mark[N];
map<pair<int, int>, int> fans;
bitset<N> atend[msq];
int n, m, q, till;
vector<pair<pair<int, int>, int> > e;
vector<int> vec;
void dfs(int v, int c) {
  mark[v][c] = 1;
  vec.push_back(v);
  for (auto u : gr[v][c])
    if (!mark[u][c]) dfs(u, c);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    x--, y--;
    gr[x][c].push_back(y), gr[y][c].push_back(x);
    e.push_back({{x, y}, c});
  }
  for (auto p : e) {
    vec.clear();
    if (!mark[p.first.first][p.second]) dfs(p.first.first, p.second);
    if (vec.size() >= sq) {
      for (auto x : vec) atend[till][x] = 1;
      till++;
    } else {
      for (int i = 0; i < vec.size(); i++)
        for (int j = i + 1; j < vec.size(); j++)
          fans[make_pair(min(vec[i], vec[j]), max(vec[i], vec[j]))]++;
    }
  }
  cin >> q;
  for (int t = 0; t < q; t++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    if (u > v) swap(u, v);
    int rtr = fans[make_pair(u, v)];
    for (int i = 0; i < till; i++)
      if (atend[i][u] == 1 && atend[i][v] == 1) rtr++;
    cout << rtr << "\n";
  }
  return 0;
}
