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

		}
	}
}