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. 
/**
 * Created by user on 12/15/2014.
 */
import java.util.
        *;
public class three
{
    public static Set<Integer>[][] adj;
    public static int[] vis;
    public static int cnt = 0;
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int m  =input.nextInt();
        vis = new int[n+1];
        adj = new Set[n+1][n+1];
        for(int i = 0; i<= n; i++)
            for(int j = 0; j <= n; j++)
                adj[i][j] = new HashSet<Integer>();
        for(int i =0; i < m; i++)
        {
            Integer a = new  Integer(input.nextInt());
            Integer b = new Integer(input.nextInt());
            Integer c= new Integer(input.nextInt());
            adj[a][b].add(c);
            adj[b][a].add(c);
        }
        int q = input.nextInt();
        for(int i = 0; i < q; i++)
        {
            Integer a = new Integer(input.nextInt());
            Integer b = new Integer(input.nextInt());
            cnt = 0;
            for(int j = 0; j <= m; j++)
            {
                setVis();
                dfs(a, b, j);
            }
            System.out.println(cnt);
        }

    }
    public static void dfs(int a, int b, Integer color)
    {
        vis[a] = 1;
        if(a == b)
            cnt++;
        for(int i = 0; i < adj[a].length; i++)
        {
            if(vis[i]!= 1 &&  adj[a][i].contains(color))
            {
               dfs(i, b, color);
            }
        }
        //return 0;
    }
    public static void setVis()
    {
        for(int i = 0; i < vis.length; i++)
            vis[i] = 0;

    }

}
