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 + 100;
int n, m, q;
vector<int> s[N];
unordered_map<int, int> par[N], res[N];
int root(int v, int c) {
  if (par[v].find(c) == par[v].end()) {
    par[v][c] = -1;
    return v;
  }
  if (par[v][c] < 0) return v;
  return (par[v][c] = root(par[v][c], c));
}
void add(int u, int v, int c) {
  if (par[v].find(c) == par[v].end()) par[v][c] = -1, s[v].push_back(c);
  if (par[u].find(c) == par[u].end()) par[u][c] = -1, s[u].push_back(c);
  if ((u = root(u, c)) == (v = root(v, c))) return;
  if (par[v][c] > par[u][c]) swap(u, v);
  par[v][c] += par[u][c];
  par[u][c] = v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  while (m--) {
    int u, v, c;
    cin >> u >> v >> c;
    add(u, v, c);
  }
  cin >> q;
  while (q--) {
    int u, v;
    cin >> u >> v;
    if (res[u].find(v) == res[u].end()) {
      if (s[v].size() > s[u].size()) swap(u, v);
      for (auto e : s[v]) {
        if (par[v].find(e) == par[v].end()) par[v][e] = v;
        if (par[u].find(e) == par[u].end()) continue;
        res[v][u] += (root(v, e) == root(u, e));
      }
    }
    res[u][v] = res[v][u];
    cout << res[v][u] << "\n";
  }
}
