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;
int n, m, q;
vector<int> H[200010];
unordered_map<int, int> G[200010], ans[200010];
int find(int x, int c) {
  if (x != G[x][c]) {
    G[x][c] = find(G[x][c], c);
    return G[x][c];
  } else
    return x;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v, c;
    scanf("%d %d %d", &u, &v, &c);
    if (!G[u].count(c)) G[u][c] = u, H[u].push_back(c);
    if (!G[v].count(c)) G[v][c] = v, H[v].push_back(c);
    int au = find(u, c), av = find(v, c);
    if (au != av) G[au][c] = av;
  }
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d %d", &u, &v);
    if (G[u].size() > G[v].size()) swap(u, v);
    if (ans[u].count(v))
      cout << ans[u][v] << endl;
    else {
      int tmp = 0;
      for (int i = 0; i < H[u].size(); i++) {
        int c = H[u][i];
        tmp += (G[u].count(c) && G[v].count(c) && find(u, c) == find(v, c));
      }
      cout << tmp << endl;
      ans[u][v] = tmp;
    }
  }
}
