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