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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.PriorityQueue;
import java.io.IOException;
import java.io.Reader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Nathaniel Lahn
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            long n = in.nextInt();
            long k = in.nextInt();

            String str = Long.toBinaryString(n);
            str = (new StringBuilder(str)).reverse().toString();

            PriorityQueue<Long> nums = new PriorityQueue<>();
            for (int i = 0; i < str.length(); i++) {
                if (str.charAt(i) == '1') {
                    long cur = (1L << (long) i);
                    nums.add(-cur);
                }

            }
//       out.println(nums);


            if (nums.size() > k) {
                out.println("NO");
                return;
            } else {
                while (nums.size() < k) {
                    long next = -nums.poll();
                    if (next == 1) {
                        out.println("NO");
                        return;
                    }

                    nums.add(-(next / 2L));
                    nums.add(-(next / 2L));
//                if (-(next/2L) == 2918) {
//                    out.println(next);
//                }
                }
            }

            StringBuilder sb = new StringBuilder();
            out.println("YES");
            for (long i : nums) {
                sb.append((-i) + " ");
            }
            out.println(sb.toString().trim());
        }

    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner(InputStream in) {
            this(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}