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 SolveA {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        for (int kases = scanner.nextInt(); kases > 0; --kases) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            if (!solve(n, k)) {
                System.out.println("NO");
            }
        }
    }

    private static boolean solve(int n, int k) {
        if (k > n)
            return false;

        int[] res = new int[k];

        if (k == 1) {
            res[0] = n;
        } else if (n % 2 == 1) {
            if (k % 2 != 1)
                return false;
            for (int i = 0; i < k; i++) {
                res[i] = 1;
            }
            res[0] += n - k;
        } else {
            // odds
            if ((n - k) % 2 == 0) {
                for (int i = 0; i < k; i++) {
                    res[i] = 1;
                }
                res[0] += n - k;
            } else if (n >= 2 * k && (n - 2 * k) % 2 == 0) {
                for (int i = 0; i < k; i++) {
                    res[i] = 2;
                }
                res[0] += n - 2 * k;
            } else {
                return false;
            }
        }

        System.out.println("YES");
        for (int i = 0; i < k; i++) {
            System.out.print(res[i] + " ");
        }
        System.out.println();
        return true;
    }
}