Problem: You are given two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

In other words, find a_1, a_2, …, a_k such that all a_i>0, n = a_1 + a_2 + … + a_k and either all a_i are even or all a_i are odd at the same time.

If such a representation does not exist, then report it.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100).

Output

For each test case print:

  * YES and the required values a_i, if the answer exists (if there are several answers, print any of them); 
  * NO if the answer does not exist. 



The letters in the words YES and NO can be printed in any case.

Example

Input


8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9


Output


YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

Solution (in JAVA):

Solution:
import java.util.*;
public class b1352
{
	public static void main(String args[])
	{
		Scanner in=new Scanner(System.in);
		int t,n,k,i,p;
		t=in.nextInt();
		while(t-->0)
		{p=0;
			n=in.nextInt();
			k=in.nextInt();
			int ans[]=new int[k];
			Arrays.fill(ans,1);
			ans[0]=n-(k-1);
			if(ans[0]%2==0&&(i=n-2*(k-1))>0&&i%2==0)
			{
				Arrays.fill(ans,2);
				ans[0]=i;
			}
			p=ans[0]%2;
			for(i=0;i<k;i++)
			{
				if(p!=ans[i]%2&&ans[i]>0)
					break;
			}
			if(i!=k)
				System.out.println("NO");
			else
			{
				System.out.println("YES");
				for(i=0;i<k;i++)
					System.out.print(ans[i]+" ");
			}
		}
	}
}