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;
map<int, int> p[100005];
int get_fa(int color, int v) {
  if (!p[color].count(v) || p[color][v] == v) return p[color][v] = v;
  return p[color][v] = get_fa(color, p[color][v]);
}
void put(int color, int a, int b) {
  a = get_fa(color, a);
  b = get_fa(color, b);
  if (a != b) {
    if (rand() & 1) swap(a, b);
    p[color][b] = a;
  }
}
vector<pair<int, int> > A[100005];
map<pair<int, int>, int> M;
int main() {
  int n, m, a, b, c;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    c--;
    put(c, a, b);
  }
  for (int color = 0; color < m; color++) {
    for (map<int, int>::iterator it = p[color].begin(); it != p[color].end();
         it++) {
      A[it->first].push_back(make_pair(color, it->second));
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < A[i].size(); j++) {
      A[i][j].second = get_fa(A[i][j].first, i);
    }
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (A[a].size() > A[b].size() || (A[a].size() == A[b].size() && a > b)) {
      swap(a, b);
    }
    if (M.count(make_pair(a, b))) {
      printf("%d\n", M[make_pair(a, b)]);
      continue;
    }
    int ans = 0;
    for (int j = 0; j < A[a].size(); j++) {
      if (binary_search(A[b].begin(), A[b].end(), A[a][j])) ans++;
    }
    M[make_pair(a, b)] = ans;
    printf("%d\n", ans);
  }
  return 0;
}
