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.*;
import javafx.util.Pair;
public class Test {
    static Scanner input = new Scanner(System.in);
    static int n,m ;
    static ArrayList< ArrayList< Pair<Integer,Integer> >> arr ;
    static boolean [][]visited ;
    static int need= 1 ;
    static int dfs(int start,int prevColor)
    {
        if(start==need) 
            return 1 ;
        int ans =0;
        for(int i=0;i<arr.get(start).size();i++)
        {
            if((prevColor==arr.get(start).get(i).getValue()||prevColor==0)&&
                    !visited[arr.get(start).get(i).getKey()][arr.get(start).get(i).getValue()])
            {
                visited[arr.get(start).get(i).getKey()][arr.get(start).get(i).getValue()]=true;
                ans+=dfs(arr.get(start).get(i).getKey(),arr.get(start).get(i).getValue());
            }
        }
        return ans ;
    }
    public static void main(String[] args) {
        n =input.nextInt();m=input.nextInt();
        arr = new ArrayList<>(n) ;
        for (int i = 0; i < n; i++) 
            arr.add(new ArrayList< Pair<Integer,Integer> >()); 
        
        for(int i = 0 ;i <m ;i++)
        {
            int v1 = input.nextInt(),v2=input.nextInt(),w=input.nextInt();
            arr.get(v1-1).add(new Pair(v2-1,w));
            arr.get(v2-1).add(new Pair(v1-1,w));
        }
        int q = input.nextInt();
        for (int i = 0; i < q; i++) {
            int start = input.nextInt()-1;
            need= input.nextInt()-1;
            visited= new boolean [n][m+1];
            System.out.println(dfs(start,0));
        }     
    }  
}