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;
vector<int> V[100000];
vector<pair<int, int>> C[100000];
bool seen[100000];
set<int> qv[100000];
set<int> seenset;
map<pair<int, int>, int> res;
void dfs(int x) {
  seen[x] = true;
  if (seenset.size() < qv[x].size()) {
    for (auto i : seenset)
      if (qv[x].find(i) != qv[x].end()) {
        int a = min(x, i), b = max(x, i);
        res[{a, b}]++;
      }
  } else {
    for (auto i : qv[x])
      if (seenset.find(i) != seenset.end()) {
        int a = min(x, i), b = max(x, i);
        res[{a, b}]++;
      }
  }
  seenset.insert(x);
  for (auto i : V[x])
    if (!seen[i]) dfs(i);
}
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);
    a--, b--, c--;
    C[c].push_back({a, b});
  }
  int Q;
  scanf("%d", &Q);
  vector<pair<int, int>> queries;
  for (int i = 0; i < Q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    qv[a].insert(b);
    qv[b].insert(a);
    queries.push_back({min(a, b), max(a, b)});
  }
  for (int i = 0; i < M; i++)
    if (C[i].size()) {
      memset(seen, 0, sizeof(seen));
      for (auto& elem : C[i]) {
        V[elem.first].push_back(elem.second);
        V[elem.second].push_back(elem.first);
      }
      for (auto& elem : C[i]) {
        int a = elem.first, b = elem.second;
        if (!seen[a]) {
          seenset.clear();
          dfs(a);
        }
      }
      for (auto& elem : C[i]) {
        V[elem.first].clear();
        V[elem.second].clear();
      }
    }
  for (auto elem : queries) printf("%d\n", res[{elem.first, elem.second}]);
  return 0;
}
