Problem: A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....

You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).

Output

If it is impossible to represent n as the sum of k powers of two, print NO.

Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.

Examples

Input


9 4


Output


YES
1 2 2 4 


Input


8 1


Output


YES
8 


Input


5 1


Output


NO


Input


3 7


Output


NO

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;
public class C {
	public static int log;
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int cur = 0;
		int n = Integer.parseInt(s.next());
		int k = Integer.parseInt(s.next());
		int[] bin = toBin(n);
		for (int i = 0; i < 31; i++)
		{
			if (bin[i] == 1)
			{
				cur++;
			}
		}
		if (k > n)
		{
			System.out.println("NO");
			return;
		}
		if (cur > k)
		{
			System.out.println("NO");
			return;
		}
		//System.out.println(Arrays.toString(bin));
		while (cur != k)
		{
			//System.out.println(Arrays.toString(bin));
			bin[log]--;
			bin[log - 1] += 2;
			if (bin[log] == 0)
			{
				log--;
			}
			cur++;
		}
		//System.out.println(Arrays.toString(bin));
		System.out.println("YES");
		for (int i = 0; i < 31; i++)
		{
			for (int j = 0; j < bin[i]; j++)
			{
				System.out.print((int) Math.pow(2, i) + " ");
			}
		}
		System.out.println();
	}
	
	public static int[] toBin(int n)
	{
		int[] ans = new int[31]; 
		
        int i = 0; 
        while (n > 0)  
        { 
            ans[i] = n % 2; 
            n = n / 2; 
            i++; 
        }
        log = i-1;
        return ans;
	}
}