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.ArrayList;
import java.util.Scanner;
public class ieee1{
    public static void main(String[] args){
        Scanner scn=new Scanner(System.in);
        int t=scn.nextInt();
        while(t-->0){
            long n=scn.nextLong();
            long k=scn.nextLong();
            if(n<k){
                System.out.println("NO");
            }
            else if(n%2!=0 && k%2==0){
                System.out.println("NO");
            }
            else if(n%2==0){
                if(n%k==0){
                    System.out.println("YES");
                    long ans=n/k;
                    while(k-->0){
                        System.out.print(ans+" ");
                    }
                    System.out.println();
                }
                else{
                    if(k<=n/2){
                        long toprint=2;
                        long upto=k-1;
                        long last=n-(2*upto);
                        System.out.println("YES");
                        while(upto-->0){
                            System.out.print(2+" ");
                        }
                        System.out.print(last);
                        System.out.println();
                    }
                    else{
                        long upto=k-1;
                        long last=n-(1*upto);
                        if(last%2!=0){
                            System.out.println("YES");
                            while(upto-->0){
                                System.out.print(1+" ");
                            }
                            System.out.print(last);
                            System.out.println();
                        }
                        else{
                            System.out.println("NO");
                        }
                    }
                }

            }
            else if(n%2!=0){
                System.out.println("YES");
                long upto=k-1;
                long last=n-(1*upto);
                while(upto-->0){
                    System.out.print(1+" ");
                }
                System.out.print(last);
                System.out.println();
            }
        }
}
    }