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;
map<int, int> f[111111];
vector<pair<int, int> > g[111111];
map<pair<int, int>, int> ans;
int F(int ind, int x) {
  if (!f[ind].count(x) || f[ind][x] == x) return f[ind][x] = x;
  return f[ind][x] = F(ind, f[ind][x]);
}
void M(int c, int a, int b) {
  a = F(c, a);
  b = F(c, b);
  if (rand() % 2) swap(a, b);
  f[c][a] = b;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    c--;
    M(c, a, b);
  }
  for (int i = 0; i < m; i++) {
    for (map<int, int>::iterator itr = f[i].begin(); itr != f[i].end(); itr++) {
      g[itr->first].push_back({i, F(i, itr->first)});
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (g[a].size() > g[b].size()) swap(a, b);
    if (ans.count({a, b})) {
      printf("%d\n", ans[{a, b}]);
      continue;
    }
    int res = 0;
    for (int i = 0; i < g[a].size(); i++) {
      if (binary_search(g[b].begin(), g[b].end(), g[a][i])) res++;
    }
    printf("%d\n", ans[{a, b}] = ans[{b, a}] = res);
  }
  return 0;
}
