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;
vector<pair<int, int> > c[100100];
vector<int> p[100100];
vector<int> h;
map<int, int> st[100100];
int a[100100], k;
bool v[100100];
void dfs(int x) {
  if (v[x]) return;
  v[x] = 1;
  a[k++] = x;
  for (int i = 0; i < p[x].size(); i++) {
    dfs(p[x][i]);
  }
}
void pro(int x) {
  k = 0;
  for (int i = 0; i < c[x].size(); i++) {
    p[c[x][i].first].push_back(c[x][i].second);
    p[c[x][i].second].push_back(c[x][i].first);
  }
  for (int i = 0; i < c[x].size(); i++) {
    if (!v[c[x][i].first]) {
      k = 0;
      dfs(c[x][i].first);
      for (int i = 0; i < k; i++) {
        for (int j = i + 1; j < k; j++) {
          st[min(a[i], a[j])][max(a[i], a[j])]++;
        }
      }
    }
  }
  for (int i = 0; i < c[x].size(); i++) {
    p[c[x][i].first].pop_back();
    p[c[x][i].second].pop_back();
    v[c[x][i].first] = 0;
    v[c[x][i].second] = 0;
  }
}
int per[100002][350];
int pr(int x, int y) {
  if (per[x][y] == x) {
    return x;
  }
  return per[x][y] = pr(per[x][y], y);
}
int main() {
  int n, m, a, b, c1;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c1);
    c[c1].push_back(make_pair(a, b));
  }
  for (int i = 1; i <= m; i++) {
    if (c[i].size() >= 340) {
      h.push_back(i);
    } else {
      pro(i);
    }
  }
  for (int i = 0; i < h.size(); i++) {
    for (int j = 0; j <= n; j++) per[j][i] = j;
  }
  for (int i = 0; i < h.size(); i++) {
    for (int j = 0; j < c[h[i]].size(); j++) {
      int x = pr(c[h[i]][j].first, i);
      int y = pr(c[h[i]][j].second, i);
      per[x][i] = y;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &a, &b);
    int r = st[min(a, b)][max(a, b)];
    for (int i = 0; i < h.size(); i++) {
      if (pr(a, i) == pr(b, i)) r++;
    }
    printf("%d\n", r);
  }
  return 0;
}
