Problem: 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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
    public static void main(String[] args) throws Exception
    {
        try
        {
            new Function().run();
        }catch(Exception e)
        {}
    }
}
class Function
{
    public Scanner cin=new Scanner(new BufferedInputStream(System.in)),fin;
    public PrintStream cout=System.out,fout;
    long []f=new long [1005];
    long inf=(long)(1e9);
    public void run()throws Exception
    {
        int i,j,n;
        long s,k;
        s=cin.nextLong();
        k=cin.nextLong();
        long []ans=new long [1000];
        f[0]=1;
        for(i=1;i<1000;i++)
        {
            for(j=i-1;j>=i-k&&j>=0;j--)
            {
                f[i]+=f[j];
            }
            if(f[i]>inf) break;
        }
        for(i=0;i<1000;i++)
        {
            if(f[i]<=s&&f[i+1]>s)
            {
                break;
            }
        }
        int cnt=0;
        for(j=i;j>=0;j--)
        {
            //cout.println(f[j]);
            if(s>=f[j])
            {
                s-=f[j];
                ans[cnt++]=f[j];
            }
            if(s==0) break;
        }
        if(cnt==1)
        {
            ans[cnt++]=0;
        }
        cout.println(cnt);
        for(i=0;i<cnt;i++)
        {
            if(i>0) cout.print(" ");
            cout.print(ans[i]);
        }
        cout.println("");
    }
}