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.Scanner;
public class SameParity 
{
	public static void main(String[] args) 
	{
		Scanner input=new Scanner(System.in);
		int T;
		T=input.nextInt();
		for(int j=0;j<T;j++)
		{
			int n,k;
			n=input.nextInt();
			k=input.nextInt();
			if(n%2==1 && k%2==0)
				System.out.println("NO");
			else 
			{
				if((n%2==0 && k%2==0) || (n%2==1 && k%2==1))
				{
					if(n > k-1)
					{
						int i;
						System.out.println("YES");
						for(i=0;i<k-1;i++)
							System.out.print("1 ");
						System.out.println(n-(k-1));
					}
					else 
						System.out.println("NO");
				}
				else if(n/2 > k-1)
				{
					int i;
					System.out.println("YES");
					for(i=0;i<k-1;i++)
						System.out.print("2 ");
					System.out.println(n-2*(k-1));
				}
				else 
					System.out.println("NO");
			}
		}
		input.close();
	}

}