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 paritySummands {

	public static void main(String[] args) {
		Scanner s = new Scanner (System.in);
		int t = s.nextInt();
		StringBuilder res = new StringBuilder();
		while (t-->0)
		{
			int n = s.nextInt();
			int k = s.nextInt();
			if (n-(k-1) > 0 && (n-(k-1))%2==1)
			{
				res.append("YES \n");
				for (int i=1;i<=k-1;i++)
					res.append(1+" ");
				res.append((n-(k-1)));
			}
			else if (n-2*(k-1) >0 && (n-2*(k-1))%2==0)
			{
				res.append("YES \n");
				for (int i=1;i<=k-1;i++)
					res.append(2+" ");
				res.append((n-2*(k-1)));
			}
			else
			{
				res.append("NO");
			}
			res.append("\n");
		}
		System.out.println(res);
	}
}