Mr. Kitayuta has just bought an undirected graph consisting of 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 ≤ 100, 1 ≤ m ≤ 100), 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 ≤ 100), 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. 
2 1
1 2 1
1
1 2
 2 3
1 2 3
1 2 2
1 2 1
1
1 2
 2 5
1 2 1
1 2 2
1 2 3
1 2 4
1 2 5
1
1 2

1
 3
 5


import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {

	public static Node[]list=new Node[101];
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int m=in.nextInt();
		for(int i=0;i<101;i++)
		{
			list[i]=new Node();
		}
		for(int i=0;i<m;i++)
		{
			int u=in.nextInt()-1;
			int v=in.nextInt()-1;
			int c=in.nextInt();
			
			list[u].colors.add(c);
			list[u].sosedi.add(v);
			
			list[v].colors.add(c);
			list[v].sosedi.add(u);
		}
		int q=in.nextInt();
		for(int i=0;i<q;i++)
		{
			int a=in.nextInt()-1;
			int b=in.nextInt()-1;
			System.out.println(BFS(a,b));
		}

	}
	public static int BFS(int a,int b)
	{
		int res=0;
		Queue<pole> q=new LinkedList<pole>();
		boolean[]visited=new boolean[101];
		boolean v[]=new boolean[101];
		Arrays.fill(v, false);
		for(int k=0;k<list[a].colors.size();k++)
		{
			for(int i=0;i<101;i++)
				visited[i]=false;
			
			if(v[list[a].colors.get(k)]) continue;
			v[list[a].colors.get(k)]=true;
			
			q.add(new pole(a,list[a].colors.get(k)));

			while(!q.isEmpty())
			{
			pole tmp=q.poll();
			if(tmp.x==b)
			{
				res++;
				break;
			}
			if(visited[tmp.x]) continue;
			visited[tmp.x]=true;
			
			for(int i=0;i<list[tmp.x].sosedi.size();i++)
			{
				if(list[tmp.x].colors.get(i)==tmp.c && !visited[list[tmp.x].sosedi.get(i)])
					q.add(new pole(list[tmp.x].sosedi.get(i),list[tmp.x].colors.get(i)));
			}
		}
		}
		return res;
	}

}
class pole
{
	public int x;
	public int c;
	public pole(int x,int c)
	{
		this.x=x;
		this.c=c;
	}
}
class Node
{
	public ArrayList<Integer> colors;
	public ArrayList<Integer> sosedi;
	public Node()
	{
		sosedi=new ArrayList<Integer>();
		colors=new ArrayList<Integer>();
	}
}
