Numbers k-bonacci (k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:

  * F(k, n) = 0, for integer n, 1 ≤ n < k; 
  * F(k, k) = 1; 
  * F(k, n) = F(k, n - 1) + F(k, n - 2) + ... + F(k, n - k), for integer n, n > k. 



Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.

You've got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers. 

Input

The first line contains two integers s and k (1 ≤ s, k ≤ 109; k > 1).

Output

In the first line print an integer m (m ≥ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a1, a2, ..., am. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.

It is guaranteed that the answer exists. If there are several possible answers, print any of them.

Examples

Input

5 2


Output

3
0 2 3


Input

21 5


Output

3
4 1 16
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author ayush
 */
import java.util.*;
public class WellKnownNumbers {
     static int a[]=new int[50];
    public static void main(String sp[])throws Exception
    {   
    	int n;
        int temp;
            for(int i=0;i<50;i++)
                a[i]=-1;
            int s,k;
        Vector v=new Vector();
        Scanner sc=new Scanner(System.in);
        Enumeration er=v.elements();
        s=sc.nextInt();
        k=sc.nextInt();

        n=k;
        while(F(k,n)<s)
        {
         n++;
        }
        //for(int i=0;i<n-k;i++)
          //  System.out.println(a[i]);
       
        temp=s;
        
        
        while(temp!=0)
        {	//System.out.println("hi");
               if(a[n-k]<=temp)
               {
                    v.add(new Integer(a[n-k]));
                    temp=temp-a[n-k];
                    n--;
                }
                else
                   n--;

        }
        v.add(new Integer(0));
        System.out.println(v.size());
        while(er.hasMoreElements())
        {
            
            System.out.print(er.nextElement()+" ");
        }


    }
    static int F(int k,int n)
    {
        if(n<k)
            return 0;
        else
            if(k==n)
            {
                    a[n-k]=1;
                return a[n-k];
            }
            else
                if(a[n-k]!=-1)
                    return(a[n-k]);
                else
                {   int sum=0;
                    for(int i=1;(i<=k)&&(n-i>=k);i++)
                        sum=sum+F(k,n-i);
                    a[n-k]=sum;
                }
        return a[n-k];
    }
}
