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 rec
{
    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();
            if(n==k)
            {
                System.out.println("Yes");
                for(int i=1;i<=k;i++)
                {
                    System.out.print(n/k+" ");
                }
                System.out.println();
            }
            else if(k>n)
            {
                System.out.println("No");
            }
            else
            {
                int rem=n%k;
                int div=n/k;
                if((rem+div)%2!=0 && div%2!=0)
                {
                    System.out.println("Yes");
                    for(int i=1;i<k;i++)
                    {
                        System.out.print(div+" ");
                    }
                    System.out.println(div+rem);
                }
                else if((rem+div)%2==0 && div%2==0)
                {
                    System.out.println("Yes");
                    for(int i=1;i<k;i++)
                    {
                        System.out.print(div+" ");
                    }
                    System.out.println(div+rem);
                }
                else
                {
                    if((k-1)%2!=0)
                    {
                        System.out.println("No");
                    }
                    else if(div==1)
                    {
                        System.out.println("No");
                    }
                    else
                    {
                        System.out.println("Yes");
                        for(int i=1;i<=k/2;i++)
                        {
                            System.out.print((div-1)+" "+(div+1)+" ");
                        }
                        System.out.println(div+rem);
                    }
                }
            }
        }
    }
}