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;
const int N = 100100;
int n, m, q;
map<int, int> fu[N];
vector<pair<int, int> > cols[N];
map<pair<int, int>, int> ans;
int fufind(int c, int i) {
  if (fu[c].find(i) == fu[c].end() || fu[c][i] == i) {
    return i;
  } else {
    return fu[c][i] = fufind(c, fu[c][i]);
  }
}
void fujoin(int c, int i, int j) {
  i = fufind(c, i);
  j = fufind(c, j);
  if (i != j) {
    fu[c][i] = j;
    fu[c][j] = j;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (m); ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    fujoin(--c, --a, --b);
  }
  for (int c = 0; c < (m); ++c)
    for (__typeof((fu[c]).begin()) i = (fu[c]).begin(); i != (fu[c]).end();
         ++i) {
      cols[i->first].push_back(make_pair(c, fufind(c, i->second)));
    }
  scanf("%d", &q);
  for (int i = 0; i < (q); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    pair<int, int> key = make_pair(a, b);
    if (ans.find(key) == ans.end()) {
      if (cols[a].size() > cols[b].size()) {
        swap(a, b);
      }
      int answer = 0;
      for (__typeof((cols[a]).begin()) i = (cols[a]).begin();
           i != (cols[a]).end(); ++i)
        if (binary_search(cols[b].begin(), cols[b].end(), *i)) {
          ++answer;
        }
      ans[key] = answer;
    }
    printf("%d\n", ans[key]);
  }
}
