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.*;
public class Main{
	static int T;
	public static void main(String[] args) throws IOException{
		StreamTokenizer re = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
		re.nextToken(); T = (int)re.nval;
		for(int p = 1;p <= T;p++){
			re.nextToken(); int n = (int)re.nval;
			re.nextToken(); int k = (int)re.nval;
			if(k > n){
				pr.println("NO");
				pr.flush();
				continue;
			}
			if((k&1) == 0 && (n&1) == 1) // k是偶数  n是奇数不行
				pr.println("NO");
			if((k&1) == 0 && (n&1) == 0){ // k n 都是偶数
				pr.println("YES");
				for(int i = 1;i < k;i++)
					pr.print(1+" ");
				pr.println(n-(k-1));
			}
			if((k&1) == 1 && (n&1) == 1){ // k n 都是奇数
				pr.println("YES");
				for(int i = 1;i < k;i++)
					pr.print(1+" ");
				pr.println(n-(k-1));
			}
			if((k&1) == 1 && (n&1) == 0){ // k是奇数  n是偶数
				if(k*2 > n)
					pr.println("NO");
				else{
					pr.println("YES");
					for(int i = 1;i < k;i++)
						pr.print(2+" ");
					pr.println(n-2*(k-1));
				}
			}
		}
		pr.flush();
	}
}