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 n, m, h, s, t, mm;
int fx, fy;
const int maxn = 100001;
int f[200001], d[maxn], cnt, x, y, z, q;
unordered_map<int, int> tree[100001];
map<pair<int, int>, int> ans;
int find(int k) {
  if (f[k] != k) f[k] = find(f[k]);
  return f[k];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= 2 * m; i++) f[i] = i;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    d[x]++;
    d[y]++;
    if (!tree[x][z]) tree[x][z] = ++cnt;
    if (!tree[y][z]) tree[y][z] = ++cnt;
    int xx = tree[x][z];
    int yy = tree[y][z];
    fx = find(f[xx]);
    fy = find(f[yy]);
    f[fy] = fx;
  }
  scanf("%d", &q);
  {
    for (int i = 1; i <= q; i++) {
      scanf("%d%d", &x, &y);
      if (d[x] > d[y]) swap(x, y);
      if (d[x] * d[x] > m && ans[pair<int, int>(x, y)]) {
        cout << ans[pair<int, int>(x, y)] - 1 << endl;
        continue;
      }
      int tem = 0;
      for (auto a = tree[x].begin(); a != tree[x].end(); a++) {
        auto b = tree[y].find(a->first);
        if (b != tree[y].end()) tem += (find(a->second) == find(b->second));
      }
      cout << tem << endl;
      if (d[x] * d[x] > m) ans[pair<int, int>(x, y)] = tem + 1;
    }
  }
  fclose(stdin);
}
