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 S1352BT{
	static int size=97;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		    int t=sc.nextInt();
            while(t-->0){
			int n=sc.nextInt();
			int k=sc.nextInt();
 
			int one=n-(k-1);
			int two=n-2*(k-1);
			if(one%2!=0 && one>0  ){
				System.out.println("YES");
				for(int i=1;i<k;i++)
				{
					System.out.print("1 ");

				}
				System.out.println(""+one);

			}
        else if(two%2==0 && two>0  ){
				System.out.println("YES");
				for(int i=1;i<k;i++)
				{
					System.out.print("2 ");

				}
				System.out.println(""+two);

			}

			
			else{
				System.out.println("NO");

			}
		}
	}
}