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.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

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

    static class C {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.ni();
            int k = in.ni();
            if (k < Integer.bitCount(n))
                out.println("NO");
            else {
                int[] b = new int[32];
                int max = 0;
                for (int i = 0; i < 32; i++) {
                    if ((n & (1 << i)) != 0) {
                        b[i] = 1;
                        k--;
                        max = Math.max(max, i);
                    }
                }
                int cur = max;
                while (cur > 0 && k > 0) {
                    if (b[cur] == 0) {
                        cur--;
                        continue;
                    }
                    b[cur]--;
                    b[cur - 1] += 2;
                    k--;
                }
                if (k > 0) {
                    out.println("NO");
                } else {
                    out.println("YES");
                    for (int i = 0; i < 32; i++) {
                        for (int j = 0; j < b[i]; j++) {
                            out.print((1 << i) + " ");
                        }
                    }
                    out.println();
                }
            }
        }

    }

    static class InputReader extends BufferedReader {
        public InputReader(InputStream st) {
            super(new InputStreamReader(st));
        }

        public String readLine() {
            try {
                return super.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        private int readByte() {
            try {
                return read();
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }

        public int ni() {
            int num = 0, b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

    }
}