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.*;

public class B {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int T = scanner.nextInt();
        for (int t = 1; t <= T; t++) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            sameParitySummands(n, k);
        }
    }

    public static void sameParitySummands(int n, int k) {
        int pn = n%2;
        int pk = k%2;
        if (pn == pk) {
            if (n < k) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
                for (int i = 0; i < k-1; i++) {
                    System.out.print(1 + " ");
                }
                System.out.println(n-k+1);
            }
        } else if (pn == 1 && pk == 0) {
            System.out.println("NO");
        } else if (pn == 0 && pk == 1) {
            if (n < 2*k) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
                for (int i = 0; i < k-1; i++) {
                    System.out.print(2 + " ");
                }
                System.out.println(n-(k-1)*2);
            }
        }
    }
}