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>
struct Edge {
  int node;
  int color;
};
std::vector<Edge> edges[1 + 100];
bool colors[1 + 100][1 + 100];
bool viz[1 + 100];
void dfs(int node, int color) {
  viz[node] = 1;
  for (int i = 0; i < edges[node].size(); ++i)
    if (viz[edges[node][i].node] == 0 && edges[node][i].color == color)
      dfs(edges[node][i].node, color);
}
int main() {
  int N, M;
  std::cin >> N >> M;
  for (int i = 1; i <= M; ++i) {
    int x, y, c;
    std::cin >> x >> y >> c;
    edges[x].push_back((Edge){y, c});
    edges[y].push_back((Edge){x, c});
    colors[x][c] = 1;
    colors[y][c] = 1;
  }
  int Q;
  std::cin >> Q;
  for (int i = 1; i <= Q; ++i) {
    int x, y;
    std::cin >> x >> y;
    int ans = 0;
    for (int c = 1; c <= M; ++c) {
      if (colors[x][c] == 1 && colors[y][c] == 1) {
        dfs(x, c);
        if (viz[y] == 1) ++ans;
        for (int k = 1; k <= N; ++k) viz[k] = 0;
      }
    }
    std::cout << ans << '\n';
  }
  return 0;
}
