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. 
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

import static java.lang.System.out;

public class Main {

    Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) {
        new Main().solve();
    }

    int a[];
    List<List<Node>> gr = new ArrayList<>();
    int k = 0;

    int used[];
    void dfs(int v, int c) {
        used[v] = 1;

        for (Node next: gr.get(v)) {
            if (used[next.v] != 0 || next.c != c) {
                continue;
            }
            dfs(next.v, c);
        }
    }

    void solve() {
        int n = scanner.nextInt(), m = scanner.nextInt();

        for (int i = 0; i < n; i++) {
            gr.add(new ArrayList<>());
        }

        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt() - 1, v = scanner.nextInt() - 1, c = scanner.nextInt() - 1;
            gr.get(u).add(new Node(v, c));
            gr.get(v).add(new Node(u, c));
        }

        int q = scanner.nextInt();

        for (int i = 0; i < q; i++) {
            int u = scanner.nextInt() - 1, v = scanner.nextInt() - 1;
            int ans = 0;
            for (int j = 0; j < m; j++) {
                used = new int[n];
                dfs(u, j);
                ans += used[v];
            }
            out.println(ans);
        }
    }

    class Node {
        int v;
        int c;

        public Node(int v, int c) {
            this.v = v;
            this.c = c;
        }
    }
}