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.*;
import java.io.*;
public class Main
{
    public static void main(String []args) throws IOException {
    	FastScanner in  = new FastScanner(System.in);
    	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
    	solve(in, out);
    	in.close();
    	out.close();
    }
    static long gcd(long a, long b){ return (b==0) ? a : gcd(b, a%b); }
    static int gcd(int a, int b){ return (b==0) ? a : gcd(b, a%b); }
    private static int fact(int n) { int ans=1; for(int i=2;i<=n;i++) ans*=i; return ans; }
    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}
	String next(){while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}
    }
    
    /********* SOLUTION STARTS HERE ************/
    
    private static void solve(FastScanner in, PrintWriter out){
        int n = in.nextInt();
        int k = in.nextInt();
        PriorityQueue<Integer> heap = new PriorityQueue<Integer>(new Comparator<Integer>(){
            public int compare(Integer a,Integer b){
                if(b>a) return 1;
                else if(b<a) return -1;
                return 0;
            }
        });
        for(int i=0;i<=30;i++){
            if(((1<<i)&n)!=0){
                heap.add(1<<i);
            }
        }
        if(heap.size()>k) {out.println("NO");return;}
        boolean res = true;
        while(heap.size()<k){
            int tmp = heap.poll();
            if(tmp==1) {res = false;break;}
            heap.add(tmp/2);
            heap.add(tmp/2);
        }
        if(!res) {out.println("NO");return;}
        out.println("YES");
        while(!heap.isEmpty()){
            out.print(heap.poll()+" ");
        }out.println();
    }
    
    /************* SOLUTION ENDS HERE **********/
}