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.Scanner;
import java.util.TreeMap;


public class CF529C {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		TreeMap<Integer, Integer> map = new TreeMap<>();
		int pow = (int) (Math.log10(n)/Math.log10(2));
		int count = 0;
		while(n>0){
			int ceilPow = pow2(pow--);
			if(ceilPow<=n){
				map.put(ceilPow, n/ceilPow);
				count+=n/ceilPow;
				n%=ceilPow;
			}
		}
		if(count>k) System.out.println("NO");
		else{
			while(count<k && map.lastKey()>1){
				int last = map.lastKey();
				if(map.get(last)>1) map.put(last, map.get(last)-1);
				else map.remove(last);
				map.put(last/2, map.containsKey(last/2) ? map.get(last/2)+2 : 2);
				count++;
			}
			if(count<k) System.out.println("NO");
			else {
				System.out.println("YES");
				StringBuilder sb = new StringBuilder();
				for(int key : map.keySet()){
					int val = map.get(key);
					for(int i=0; i<val; i++) sb.append(key).append(" ");
				}
				System.out.println(sb.toString().trim());
			}
		}
	}
	
	static int pow2(int k){
		int val = 1;
		while(k-->0) val*=2;
		return val;
	}
}