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;
#pragma comment(linker, "/STACK:30000000");
const int MX = 1e5 + 5;
int n, m;
vector<pair<int, int>> c;
vector<pair<int, int>> g[MX];
int color = 1;
int used[MX];
int q;
vector<int> l[MX];
int now;
map<pair<int, int>, int> ans;
void dfs(int num, int tick) {
  if (used[num] == color) return;
  used[num] = color;
  l[num].push_back(tick);
  while (g[num].size() && g[num].back().first == now) {
    dfs(g[num].back().second, tick);
    g[num].pop_back();
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, cc;
    cin >> a >> b >> cc;
    g[a].push_back(make_pair(cc, b));
    g[b].push_back(make_pair(cc, a));
    c.push_back(make_pair(cc, a));
    c.push_back(make_pair(cc, b));
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    reverse(g[i].begin(), g[i].end());
  }
  sort(c.begin(), c.end());
  for (int i = 0; i < c.size(); i++) {
    if (i && c[i].first != c[i - 1].first) color++;
    now = c[i].first;
    if (used[c[i].second] != color) dfs(c[i].second, i);
  }
  for (int i = 1; i <= n; i++) sort(l[i].begin(), l[i].end());
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    map<pair<int, int>, int>::iterator it = ans.find(make_pair(a, b));
    if (it == ans.end()) {
      int cnt = 0;
      int pa = 0, pb = 0;
      int ta = a, tb = b;
      if (l[ta].size() > l[tb].size()) swap(ta, tb);
      for (int i = 0; i < l[ta].size(); i++) {
        if (binary_search(l[tb].begin(), l[tb].end(), l[ta][i])) cnt++;
      }
      ans[make_pair(a, b)] = cnt;
      cout << cnt << endl;
    } else {
      cout << it->second << endl;
    }
  }
  return 0;
}
