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 SameParitySummands {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while(T-->0){
			long n=sc.nextLong();
			long k=sc.nextLong();
			long k_copy=k;
			
			if((n-k+1)%2==1 && n-k_copy+1>0){
				System.out.println("YES");
				while(--k>0)
					System.out.print(1+" ");
				System.out.println(n-k_copy+1);
			}
			else if((n-(k-1)*2)%2==0 && n-(k_copy-1)*2>0){
				System.out.println("YES");
				while(--k>0)
					System.out.print(2+" ");
				System.out.println(n-(k_copy-1)*2);
			}
			else{
				System.out.println("NO");
			}
 
		}
	}

}