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;
typedef struct {
  vector<int> arcs;
} edge;
edge g[200][200];
int func(int n, int x, int y, vector<int>& colores) {
  int cant = 0;
  for (unsigned i = 0; i < colores.size(); i++) {
    vector<vector<int> > ccs(n);
    for (int j = 0; j < n; j++) ccs[j].push_back(j + 1);
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= n; k++)
        if (find(g[j][k].arcs.begin(), g[j][k].arcs.end(), colores[i]) !=
            g[j][k].arcs.end()) {
          int v1;
          int v2;
          int a = 0;
          while (find(ccs[a].begin(), ccs[a].end(), j) == ccs[a].end()) a++;
          v1 = a;
          a = 0;
          while (find(ccs[a].begin(), ccs[a].end(), k) == ccs[a].end()) a++;
          v2 = a;
          if (v1 != v2)
            while (!ccs[v2].empty()) {
              ccs[v1].push_back(ccs[v2][0]);
              ccs[v2].erase(ccs[v2].begin());
            }
        }
    }
    bool b = false;
    for (unsigned k = 0; k < ccs.size() && !b; k++)
      b = find(ccs[k].begin(), ccs[k].end(), x) != ccs[k].end() &&
          find(ccs[k].begin(), ccs[k].end(), y) != ccs[k].end();
    cant += (int)(b);
  }
  return cant;
}
int main() {
  int n, m, x, y, c;
  cin >> n >> m;
  vector<vector<int> > colores(n + 2);
  while (m--) {
    cin >> x >> y >> c;
    g[x][y].arcs.push_back(c);
    g[y][x].arcs.push_back(c);
    if (find(colores[x].begin(), colores[x].end(), c) == colores[x].end())
      colores[x].push_back(c);
    if (find(colores[y].begin(), colores[y].end(), c) == colores[y].end())
      colores[y].push_back(c);
  }
  cin >> c;
  while (c--) {
    cin >> x >> y;
    if (colores[x].size() < colores[y].size())
      cout << func(n, x, y, colores[x]) << endl;
    else
      cout << func(n, x, y, colores[y]) << endl;
  }
}
