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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem529C {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter (new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int n=s.nextInt();
		int k=s.nextInt();
		String Binary=Integer.toBinaryString(n);
		int[] arr=new int[Binary.length()];
		int countones=0;
		for(int i=Binary.length()-1;i>=0;i--) {
			arr[i]=Binary.charAt(i)-'0';
			if(arr[i]!=0) {
				countones++;
			}
		}
		int[] codes=new int[Binary.length()];
		for(int i=Binary.length()-1;i>=0;i--) {
			codes[Binary.length()-i-1]=1<<i;
//			if(arr[Binary.length()-i-1]!=0) {
//				out.println(codes[i]);
//			}
		}
		if(k>n || k<countones) {
			out.println("NO");
			out.close();
			return;
		}
		k-=countones;
		int currentpos=0;
		while(k>0) {
			if(arr[currentpos]==0) {
				currentpos++;
			}
			if(arr[currentpos]>0) {
				arr[currentpos]--;
				k--;
				arr[currentpos+1]+=2;
			}
		}
		out.println("YES");
		for(int i=0;i<Binary.length();i++) {
			if(arr[i]>0) {
				for(int j=0;j<arr[i];j++) {
					out.print(codes[i]+" ");
				}
			}
		}
		out.close();
	}

	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

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

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

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

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

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

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