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_MAX = 100000;
const int M_MAX = 100000;
const int Q_MAX = 100000;
int n, m;
struct Edge {
  int u, v;
};
vector<Edge> edges[M_MAX + 2];
int q;
int bucket;
int root[N_MAX + 2];
int findRoot(int u) {
  if (root[u] == u) return u;
  return root[u] = findRoot(root[u]);
}
void join(int u, int v) {
  u = findRoot(u);
  v = findRoot(v);
  if (u == v) return;
  root[u] = v;
}
vector<int> subNodes[N_MAX + 2];
vector<vector<int>> components;
vector<int> vcomp[N_MAX + 2];
vector<vector<int>> bigAnswers;
int id[N_MAX + 2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    Edge e;
    int c;
    cin >> e.u >> e.v >> c;
    edges[c].push_back(e);
  }
  cin >> q;
  bucket = m / sqrt(q);
  for (int i = 1; i <= n; i++) root[i] = i;
  for (int c = 1; c <= m; c++) {
    set<int> adj;
    for (Edge e : edges[c]) {
      join(e.u, e.v);
      adj.insert(e.u);
      adj.insert(e.v);
    }
    for (int u : adj) subNodes[findRoot(u)].push_back(u);
    for (int u : adj) {
      if (root[u] == u) {
        components.push_back(subNodes[u]);
        subNodes[u].clear();
      }
      root[u] = u;
    }
  }
  for (int i = 0; i < (int)components.size(); i++)
    for (int u : components[i]) vcomp[u].push_back(i);
  for (int i = 1; i <= n; i++)
    if ((int)vcomp[i].size() > bucket) {
      id[i] = (int)bigAnswers.size();
      bigAnswers.push_back(vector<int>(n + 1));
      for (int c : vcomp[i])
        for (int u : components[c]) bigAnswers.back()[u]++;
    } else
      id[i] = -1;
  for (int i = 1; i <= q; i++) {
    int u, v;
    cin >> u >> v;
    if (id[u] != -1) {
      cout << bigAnswers[id[u]][v] << "\n";
      continue;
    }
    if (id[v] != -1) {
      cout << bigAnswers[id[v]][u] << "\n";
      continue;
    }
    int ptu = 0, ptv = 0;
    int answer = 0;
    while (ptu < (int)vcomp[u].size() && ptv < (int)vcomp[v].size()) {
      if (vcomp[u][ptu] == vcomp[v][ptv]) {
        answer++;
        ptu++;
        ptv++;
      } else if (vcomp[u][ptu] < vcomp[v][ptv])
        ptu++;
      else
        ptv++;
    }
    cout << answer << "\n";
  }
  return 0;
}
