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.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class Dummy {

    private static Scanner scanner;

    public static void main(String[] args) {

        solve();
    }

    private static void solve() {
        scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();

        List<Map<Integer, List<Integer>>> data = new ArrayList<>();
        data.add(new HashMap<>()); // dummy data
        for (int i = 1; i <= m; i++) {
            Map<Integer, List<Integer>> map = new HashMap<>();
            for (int j = 1; j <= n; j++)
                map.put(j, new ArrayList<>());
            data.add(map);
        }

        for (int i = 1; i <= m; i++) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            int c = scanner.nextInt();
            data.get(c).get(a).add(b);
            data.get(c).get(b).add(a);
        }

        List<List<Set<Integer>>> solution = new ArrayList<>();
        for (int i = 0; i <= m; i++) {
            solution.add(new ArrayList<>());
        }

        for (int i = 1; i <= m; i++) {
            boolean[] visited = new boolean[n + 1];
            for (int j = 1; j <= n; j++) {
                Set<Integer> set = new HashSet<>();
                DFS(j, i, data, visited, set);
                solution.get(m).add(set);
            }
        }

        int q = scanner.nextInt();
        for (int i = 0; i < q; i++) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            int count = 0;
            for (int j = 1; j <= m; j++) {
                List<Set<Integer>> list = solution.get(j);
                for (int k = 0; k < list.size(); k++) {
                    if (list.get(k).contains(a) && list.get(k).contains(b))
                        count++;
                }
            }
            System.out.println(count);
        }
    }

    private static void DFS(int curr,
                            int color,
                            List<Map<Integer,
                            List<Integer>>> data,
                            boolean[] visited,
                            Set<Integer> set) {

        if (visited[curr])
            return;

        visited[curr] = true;
        set.add(curr);

        List<Integer> neighbours = data.get(color).get(curr);
        for (int i = 0; i < neighbours.size(); i++) {
            DFS(neighbours.get(i), color, data, visited, set);
        }
    }
}