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 Solgrid{
    private static void findit(long n,int k) {
	
	ArrayList<Integer> arr=new ArrayList<Integer>();
	if((n<k)  || (k%2==0 && n%2==1))
	{
		System.out.println("NO");
		return;
	}
	int f1=1;
	if(n%2!=k%2 && n>=(2*k))
		f1=0;
	else if(n%2!=k%2)
	{
		System.out.println("NO");
		return;
	}
	long sum=0;
	for(int i=0;i<k-1;i++)
	{
		if(f1==1)
		{
			arr.add(1);
			sum=sum+1;
		}
		else
		{
			arr.add(2);
			sum=sum+2;
		}
	}
	arr.add((int)n-(int)sum);
	
	System.out.println("YES");
	for(int i=arr.size()-1;i>=0;i--)
	{
		System.out.print(arr.get(i)+" ");
	}
	System.out.println();

        
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
	int t=sc.nextInt();
	for(int i=0;i<t;i++)
	{
		long n=sc.nextLong();
		int k=sc.nextInt();
		findit(n,k);
	}

    }
}