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 f[100100];
vector<pair<int, int> > E[100100];
vector<int> heavy;
map<pair<int, int>, int> ans;
int comp[290][100100];
vector<int> L[100100];
vector<int> feliz;
int foi[100100];
void dfs(int x, int y, int u, int cmp) {
  if (foi[x] == y) return;
  comp[u][x] = cmp;
  foi[x] = y;
  for (vector<int>::iterator it = L[x].begin(); it < L[x].end(); it++)
    dfs(*it, y, u, cmp);
}
void dfs2(int x, int y) {
  if (foi[x] == y) return;
  ;
  feliz.push_back(x);
  foi[x] = y;
  for (vector<int>::iterator it = L[x].begin(); it < L[x].end(); it++)
    dfs2(*it, y);
}
int main() {
  int n, m;
  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--;
    f[c]++;
    E[c].push_back(pair<int, int>(a, b));
  }
  int sq = 350;
  int cur = 1;
  int conta = 0;
  for (int i = 0; i < m; i++) {
    ;
    if (f[i] == 0) continue;
    if (f[i] > sq) {
      ;
      heavy.push_back(i);
      for (int j = 0; j < n; j++) L[j].clear();
      for (int j = 0; j < f[i]; j++) {
        L[E[i][j].first].push_back(E[i][j].second);
        L[E[i][j].second].push_back(E[i][j].first);
      }
      int k = 0;
      cur++;
      for (int j = 0; j < n; j++)
        if (foi[j] != cur) {
          dfs(j, cur, conta, k);
          k++;
        }
      conta++;
      continue;
    };
    for (int j = 0; j < f[i]; j++) {
      L[E[i][j].first].clear();
      L[E[i][j].second].clear();
    }
    for (int j = 0; j < f[i]; j++) {
      L[E[i][j].first].push_back(E[i][j].second);
      L[E[i][j].second].push_back(E[i][j].first);
      ;
    }
    cur++;
    feliz.clear();
    for (int j = 0; j < f[i]; j++) {
      if (foi[E[i][j].first] != cur) {
        feliz.clear();
        ;
        dfs2(E[i][j].first, cur);
        for (int a = 0; a < feliz.size(); a++)
          for (int aa = a + 1; aa < feliz.size(); aa++) {
            ans[pair<int, int>(min(feliz[a], feliz[aa]),
                               max(feliz[a], feliz[aa]))]++;
            ;
          }
      }
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b;
    scanf("%d%d", &a, &b), a--, b--;
    if (a > b) swap(a, b);
    int r = ans[pair<int, int>(a, b)];
    for (int i = 0; i < conta; i++)
      if (comp[i][a] == comp[i][b]) r++;
    printf("%d\n", r);
  }
}
