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;
set<pair<int, int> > a[100005];
set<int> ans[100005], C[100005];
map<int, int> stay[100005];
set<int>::iterator p;
int vis[100005], n, m, x, y, z, i, col, Q, number, ID;
void DFS(int x) {
  if (vis[x] == col) return;
  vis[x] = col;
  ans[x].insert(ID);
  for (set<pair<int, int> >::iterator it = a[x].begin(); it != a[x].end(); it++)
    if ((*it).first == col)
      DFS((*it).second);
    else
      break;
  while (!a[x].empty()) {
    set<pair<int, int> >::iterator it = a[x].begin();
    if ((*it).first > col) return;
    a[x].erase(it);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++)
    scanf("%d%d%d", &x, &y, &z), a[x].insert(make_pair(z, y)),
        a[y].insert(make_pair(z, x)), C[z].insert(x), C[z].insert(y);
  for (col = 1; col <= 1e5; col++)
    if (!C[col].empty())
      for (p = C[col].begin(); p != C[col].end(); p++) ++ID, DFS(*p);
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &x, &y);
    number = 0;
    if (ans[x].size() > ans[y].size()) swap(x, y);
    if (stay[x].find(y) != stay[x].end()) {
      printf("%d\n", stay[x][y]);
      continue;
    }
    for (p = ans[x].begin(); p != ans[x].end(); p++)
      if (ans[y].find(*p) != ans[y].end()) number++;
    printf("%d\n", stay[x][y] = number);
  }
}
