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;
void dfs(int u, int y, vector<set<int>> &visit,
         vector<vector<pair<int, int>>> &adj, int *coun, int color) {
  if (color > 0) visit[u].insert(color);
  int vector, c, i;
  if (u == y) {
    (*coun)++;
    return;
  }
  for (i = 0; i < adj[u].size(); i++) {
    pair<int, int> e = adj[u][i];
    vector = e.first, c = e.second;
    if (c == color || color == -1)
      if (visit[vector].find(c) == visit[vector].end()) {
        dfs(vector, y, visit, adj, coun, c);
      }
  }
  return;
}
int main() {
  int x, y, c, i, q;
  cin >> n >> m;
  vector<vector<pair<int, int>>> adj(n);
  for (i = 0; i < m; i++)
    scanf(" %d", &x), scanf(" %d", &y), scanf(" %d", &c),
        adj[x - 1].push_back({y - 1, c}), adj[y - 1].push_back({x - 1, c});
  cin >> q;
  while (q--) {
    scanf(" %d", &x), scanf(" %d", &y);
    vector<set<int>> visit(n);
    int count = 0;
    dfs(x - 1, y - 1, visit, adj, &count, -1);
    printf("%d\n", count);
  }
  return 0;
}
