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 Parity_Summands {
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);

        long t = in.nextLong();
        while(t-->0){
            long n =in.nextLong();
            long k = in.nextLong();
            boolean flagodd=false,flageven=false;
            if(((n-(k-1)*1)%2==1) && (n-(k-1)*1)>0 )
                flagodd = true;
            else if(((n-(k-1)*2)%2==0) && (n-(k-1)*2)>0)
                flageven=true;

            if(flagodd==false && flageven==false)
                System.out.println("NO");
            else if(flagodd==true){
                System.out.println("YES");
                for(int i=1;i<=k-1;i++){
                    System.out.print("1"+" ");
                }
                System.out.print((n-(k-1)*1));
                System.out.println();
            }
            else{
                System.out.println("YES");
                for(int i=1;i<=k-1;i++){
                    System.out.print("2"+" ");
                }
                System.out.print((n-(k-1)*2));
                System.out.println();
            }
        }
    }
}