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 long long MAX_N = 100 * 1000 + 24, INF = (1 << 30);
int n, m, q, a, b, c, ans, h;
map<int, vector<int>> G[MAX_N];
map<int, int> mm[MAX_N];
map<pair<int, int>, int> out;
bool mark[MAX_N];
set<int> cl[MAX_N], cs;
void dfs(int v, int k, int c) {
  mark[v] = true;
  mm[v][c] = k;
  if (G[v].count(c) > 0) {
    for (int u : G[v][c]) {
      if (mark[u] == false) {
        dfs(u, k, c);
      }
    }
  }
}
int main() {
  cin.tie(0), cout.tie(0), ios::sync_with_stdio(false);
  scanf("%d %d", &n, &m);
  while (m--) {
    scanf("%d %d %d", &a, &b, &c);
    G[a][c].push_back(b);
    G[b][c].push_back(a);
    cl[c].insert(a);
    cs.insert(c);
  }
  for (set<int>::iterator i = cs.begin(); i != cs.end(); i++) {
    memset(mark, 0, sizeof mark);
    for (set<int>::iterator j = cl[*i].begin(); j != cl[*i].end(); j++) {
      if (mark[*j] == false) {
        dfs(*j, *j, *i);
      }
    }
  }
  scanf("%d", &q);
  while (q--) {
    ans = 0;
    scanf("%d %d", &a, &b);
    if (mm[a].size() > mm[b].size()) swap(a, b);
    if (out.count(make_pair(a, b)) == 0) {
      for (pair<int, int> u : mm[a]) {
        if (mm[b].count(u.first) > 0 and u.second == mm[b][u.first]) ans++;
      }
      out[make_pair(a, b)] = ans;
      printf("%d\n", ans);
    } else
      printf("%d\n", out[make_pair(a, b)]);
  }
}
