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 Java_640_B {
    public static void main(String[] args) {
      int t,n,k;
        Scanner sc=new Scanner(System.in);
        t=sc.nextInt();
        for(int i=0;i<t;i++)
        {
        n=sc.nextInt();
        k=sc.nextInt();
        
        if(n%2==0)
        {
        Even(n,k);
        }
        else
        {
        Odd(n,k);
        }
        }
    }
    
    public static void Even(int n,int k)
    {
        int A[]=new int[k];
     if(k<=n/2)
     {
         System.out.println("Yes");
         A[0]=n-(2*(k-1));
      for(int i=1;i<k;i++)
      {
      A[i]=2;
      }
      for(int j=0;j<k;j++)
      {
          System.out.println(A[j]);
      }
     }
     else
     {
         if(k<=n && k%2==0)
         {
             System.out.println("Yes");
             A[0]=(n-k)+1;
             for(int i=1;i<k;i++)
             {
             A[i]=1;
             }
             for(int j=0;j<k;j++)
             {
                 System.out.println(A[j]);
             }
         }
         else
         {
         System.out.println("No");
         }
     }
    }
    
    public static void Odd(int n,int k)
    { 
        int A[]=new int[k];
    if(k<=n && k%2==1)
    {
        System.out.println("Yes");
     A[0]=n-(k-1);
     for(int i=1;i<k;i++)
     {
     A[i]=1;
     }
     for(int j=0;j<k;j++)
      {
          System.out.println(A[j]);
      }
    }
    else
    {
        System.out.println("No");
    }
    }
    
}