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, Ed, Q, Par[100005];
map<pair<int, int>, int> M;
vector<pair<int, int>> C[100005], E[100005];
int find(int n) { return Par[n] == n ? n : Par[n] = find(Par[n]); }
void join(int a, int b) { Par[find(a)] = find(b); }
void process(int c) {
  for (pair<int, int> p : E[c])
    Par[p.first] = p.first, Par[p.second] = p.second;
  for (pair<int, int> p : E[c]) join(p.first, p.second);
  for (pair<int, int> p : E[c]) {
    if (C[p.first].empty() || C[p.first].back().first != c)
      C[p.first].push_back({c, find(p.first)});
    if (C[p.second].empty() || C[p.second].back().first != c)
      C[p.second].push_back({c, find(p.second)});
  }
  for (pair<int, int> p : E[c])
    Par[p.first] = p.first, Par[p.second] = p.second;
}
int query(int a, int b) {
  if (((int)C[a].size()) > ((int)C[b].size())) swap(a, b);
  if (M.find({a, b}) != M.end()) return M[{a, b}];
  int ret = 0;
  for (pair<int, int> p : C[a]) {
    auto it = lower_bound(C[b].begin(), C[b].end(), p);
    if (it != C[b].end() && *it == p) ret++;
  }
  return M[{a, b}] = ret;
}
int main() {
  scanf("%d%d", &N, &Ed);
  for (int i = (1); i <= (Ed); i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    E[c].push_back({a, b});
  }
  for (int c = (1); c <= (Ed); c++) process(c);
  for (int c = (1); c <= (Ed); c++) sort(C[c].begin(), C[c].end());
  cin >> Q;
  while (Q--) {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", query(a, b));
  }
}
