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.io.*;
import java.math.*;
import java.math.BigInteger;
import java.util.*;
public class Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		for(int t=sc.nextInt();t>0;t--) {
			int m=sc.nextInt();
			int n=sc.nextInt();
			int x=0,y=0;
			x=m-(n-1);
			y=m-2*(n-1);
			if(x%2==1&&x>0) {
				System.out.println("YES");
				for(int i=0;i<n-1;i++) {
					System.out.print(1+" ");
				}
				System.out.println(x);
			}else if(y%2==0&&y>0) {
				System.out.println("YES");
				for(int i=0;i<n-1;i++) {
					System.out.print(2+" ");
				}
				System.out.println(y);
			}else {
				System.out.println("NO");
			}
		}
	}
}