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 ques1 {
 
	public static void main(String args[]) {
		Scanner s = new Scanner(System.in);
		
		int t=s.nextInt();
		int n;
		int k;
		for(int i =0;i<t;i++) {
			//System.out.println("i "+i);
			n=s.nextInt();
			k=s.nextInt();
			if(n<k) {
				System.out.println("NO");
			}
			else {
				//System.out.println(n-(k-1)%2);
			if((n-(k-1))%2==1) {
				System.out.println("YES");
				for(int j=0;j<k-1;j++) {
					System.out.print("1 ");
				}
				System.out.println(n-(k-1));
			}
			else {
				if(n-(k-1)-(k-1)>0) {
				if((n-(k-1)-(k-1))%2==0){
					System.out.println("YES");
					for(int j=0;j<k-1;j++) {
						System.out.print("2 ");
					}
					System.out.println(n-(k-1)-(k-1));
				}
				else {
					System.out.println("NO");
				}
				}
				else {
					System.out.println("NO");
				}
			}
			}
		}
	}
}