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.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;


public class Main{
	static int n,m,ans,t,x2,x1,y1,y2;
	static String s;
	static double pi=3.1415926;
	static int [][]f=new int[105][105];
	public static int get(int x,int k){
		if(f[x][k]==x)
			return x;
		else return f[x][k]=get(f[x][k], k);
	}

	public static void main(String[] args){
		InputReader in=new InputReader(System.in);
		//Scanner in=new Scanner(new BufferedInputStream(System.in));
		n=in.nextInt();
		m=in.nextInt();
		
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
				f[i][j]=i;
		for(int i=1;i<=m;i++){
			x1=in.nextInt();
			x2=in.nextInt();
			y1=in.nextInt();
			int a=get(x1, y1);
			int b=get(x2, y1);
//			System.out.println("a="+a+" b="+b);
			if(a!=b)
				f[a][y1]=b;
//			for(int ii=1;ii<=n;ii++)
//				System.out.print(" f"+ii+"="+f[ii][y1]);
//			System.out.println();
		
		}
		t=in.nextInt();
		for(int i=1;i<=t;i++){
			x1=in.nextInt();
			x2=in.nextInt();
			ans=0;
			for(int j=1;j<=m;j++){
//				System.out.println("j="+j);
//				System.out.print("x="+f(x1, j));
//				System.out.println(" y="+f(x2, j));
				if(get(x1, j)==get(x2, j))
					ans++;
			}
				
			System.out.println(ans);
		}
	}
	public static class InputReader{
		BufferedReader reader;
		StringTokenizer tokenizer;
		
		public InputReader(InputStream stream){
			reader=new BufferedReader(new InputStreamReader(System.in),32768);
			tokenizer=null;
		}
		public String next(){
			
			while (tokenizer==null||!tokenizer.hasMoreTokens()) {
				
				try {
					tokenizer=new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					throw new RuntimeException(e);
				}
				
			}
			return tokenizer.nextToken();
		}
		public int nextInt(){
			return Integer.parseInt(next());
		}
		public boolean ready() throws IOException{
			return reader.ready();
		}
	}


}
		 	   		 		 	   	    	 	    		