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 q2 {

	public static void main(String[] args) {

		Scanner a = new Scanner(System.in);
		int t = a.nextInt();
		for(int i=0;i<t;i++)
		{
			int n = a.nextInt();
			int k = a.nextInt();
			if(k>n)
			{
				System.out.println("NO");
			}
			else
			{
			int ans = n/k;
			int rem = n%k;
			if(rem%2==0)
			{
				System.out.println("YES");
				System.out.print((ans+rem) +" ");
				for(int j=0;j<k-1;j++)
				{
					System.out.print(ans +" ");
				}
			}
			else
			{
					if(k%2!=0 && ans!=1)
					{
						System.out.println("YES");
						System.out.print((ans+k+rem-1) +" ");
						for(int j=0;j<k-1;j++)
						{
							System.out.print((ans-1) +" ");
						}
					}
					else
					{
						System.out.println("NO");
					}
			}
			System.out.println();
			}
		}
	}

}