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.io.*;
import java.util.*;
public class Main{
	public static void main(String args[]) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		String o[] = new String[t];
		for(int i=0;i<t;i++){
			String in = br.readLine();
			String f[] = in.split(" ");
			int n = Integer.parseInt(f[0]);
			int k = Integer.parseInt(f[1]);

			String ans = "NO";
			int x = n-(k-1);
			int y = n-2*(k-1);

			if(x > 0 && x%2!=0){
				ans = "YES\n";
				for(int l=0;l<k-1;l++)
					ans+="1 ";
				ans+=x;
			}
			if(y > 0 && y%2==0){
				ans = "YES\n";
				for(int l=0;l<k-1;l++)
					ans+="2 ";
				ans+=y;
			}

			o[i] = ans;
		}
		for(int i=0;i<t;i++){
			System.out.println(o[i]);
		}
	}
}