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.*;
public class ProblemC {
    public static void main(String[] argc)
    {
        Scanner sc = new Scanner(System.in);
        int n, k;
        n = sc.nextInt();
        k = sc.nextInt();
        int cur = n;
        int count = 0;
        StringBuilder sb = new StringBuilder();

        int power = 1;
        Queue<Integer> candidate_q = new ArrayDeque<>();
        int candidate = 0;
        // System.out.println(cur);

        while(cur > 0)
        {
            int rem = cur % 2;
            if(rem != 0) {
                candidate = rem * power;
                // System.out.println(rem + " " + candidate + " ");
                candidate_q.offer(candidate);
            }

            cur /= 2;
            power *= 2;
        }

        // Collections.reverse(candidate_q);
        // System.out.println(candidate_List.size());

        //for(int i = 0; i < candidate_List.size(); i++)
        //    System.out.println(candidate_List.get(i));

        
        if(candidate_q.size() > k)
        {
            System.out.println("NO");
            return;
        }
        else
        {
            int loop_count = 0;
            while(candidate_q.size() != k)
            {
                int tmp = candidate_q.poll();
                if(tmp > 1)
                {
                    candidate_q.offer(tmp/2);
                    candidate_q.offer(tmp/2);
                }
                else
                {
                    loop_count++;
                    candidate_q.offer(tmp);
                }
                if(loop_count > 2000000)
                    break;
            }

            if(candidate_q.size() < k) {
                System.out.println("NO");
                return;
            }

            List<Integer> res_List = new ArrayList<>();
            while(!candidate_q.isEmpty())
            {
                res_List.add(candidate_q.poll());
            }
            sb.append("YES \n");
            Collections.sort(res_List);
            for(int i = 0; i < res_List.size(); i++)
            {
                sb.append(res_List.get(i) + " ");
            }
            System.out.println(sb.toString());
        }

    }
}