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.HashSet;
import java.util.Scanner;

public class WorkFile {
    public static boolean dfs(ArrayList<int[]>[] graph, int color,
                              ArrayList<Integer> visited, int item, int end) {
        for (int[] elem:graph[item]) {
            if (elem[1]==color && !visited.contains(elem[0])) {
                if (elem[0]==end) return true;
                ArrayList<Integer> newvisited = new ArrayList<>(visited);
                newvisited.add(elem[0]);
                if (dfs(graph, color, newvisited, elem[0], end)) {
                    return true;
                }
            }
        }
        return false;
    }
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), m = s.nextInt();
        ArrayList[] graph = new ArrayList[n];
        for (int i=0; i<n; i++) {
            graph[i] = new ArrayList<int[]>();
        }
        HashSet<Integer> set = new HashSet<>();
        for (int i=0; i<m; i++) {
            int a = s.nextInt()-1, b = s.nextInt()-1, c = s.nextInt();
            set.add(c);
            graph[a].add(new int[]{b,c});
            graph[b].add(new int[]{a,c});
        }
        int q = s.nextInt();
        StringBuilder result = new StringBuilder();
        while (q-->0) {
            int x = s.nextInt()-1, y = s.nextInt()-1;
            int res = 0;
            for (int color:set) {
                ArrayList<Integer> visited = new ArrayList();
                visited.add(x);
                if (dfs(graph, color, visited, x, y)) res++;
            }
            result.append(res).append("\n");
        }
        System.out.println(result);
    }
}