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:
//package CF529.PowersOfTwo;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        ArrayList<Integer> pow2 = new ArrayList<>();
        pow2.add(1);
        if(k>n){
            System.out.println("NO");
        }
        else {
            while (pow2.get(pow2.size() - 1) < n) {
                pow2.add(pow2.get(pow2.size() - 1) * 2);
            }
            if (pow2.get(pow2.size() - 1) > n) {
                pow2.remove(pow2.size() - 1);
            }
            int clone = n;
            int curindex = pow2.size()-1;
            LinkedList<Integer> actual =new LinkedList<Integer>();
            while(clone>0 && curindex>=0){
                if(pow2.get(curindex)<=clone) {
                    actual.add(pow2.get(curindex));
                    clone -= pow2.get(curindex);
                }
                    curindex--;
            }
            if(k<actual.size()){
                System.out.println("NO");
            }
            else{
                System.out.println("YES");
                while(actual.size()!=k){
                    int l  = actual.pollFirst();
                    if(l>1){
                        actual.addFirst(l/2);
                        actual.addFirst(l/2);
                    }
                    else{
                        actual.addLast(1);

                    }
                }
                Collections.sort(actual);
                while(actual.size()>0){
                    System.out.print(actual.pollFirst()+" ");
                }
            }

        }
    }
}