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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

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

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String[] args) {
        FastReader in = new FastReader();
        int n = in.nextInt();
        int k = in.nextInt();
        if (n < k || countBit(n) > k) {
            System.out.println("NO");
        } else {
            int[] bits = new int[32];
            System.out.println("YES");
            for (int i = 31; i >= 0; i--)
                if ((n & (1 << i)) != 0) {
                    bits[i]++;
                    k--;
                }
            StringBuilder sb = new StringBuilder();
            for (int i = 31; i >= 0; i--) {
                if (k != 0) {
                    int shift = Math.min(bits[i], k);
                    bits[i - 1] += shift * 2;
                    bits[i] -= shift;
                    k -= shift;
                }
                while (bits[i]-- > 0) {
                    sb.append(1 << i).append(" ");
                }
            }
            System.out.println(sb.toString().trim());
        }
    }

    static int countBit(int x) {
        int count = 0;
        while (x != 0) {
            count++;
            x &= (x - 1);
        }
        return count;
    }
}