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 Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t>0){
			t--;
			int n=s.nextInt();
			int k=s.nextInt();
			int l1=n-(k-1);
			int l2=n-(k-1)*2;
			if(l1%2!=0 && l1>0) {
					System.out.println("YES");
					System.out.print(l1+" ");
					for(int i=0;i<k-1;i++) {
						System.out.print(1+" ");
					}
					System.out.println();
				}
			else if(l2>0) {
				if(l2%2==0) {
					System.out.println("YES");
					System.out.print(l2+" ");
					for(int i=0;i<k-1;i++) {
						System.out.print(2+" ");
					}
					System.out.println();
				}
				else {
					System.out.println("NO");
				}
			}
			else {
				System.out.println("NO");
			}
			
		}
		
		}
	}