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.util.*;

public class Main{

  public void run(){
    Scanner in = new Scanner(System.in);
    int s = in.nextInt();
    int k = in.nextInt();
    
    int f[] = new int[100000];
    int i = 1;
    f[0] = 1;
    int sum = 1;
    while(true){
      f[i]+=sum;
      if (f[i] >= s) break;
      if (i >= k) sum-=f[i-k];
      sum+=f[i];
      i++;
    }
    /*
    for(int j=0;f[j]!=0;j++)
      System.out.print(f[j] + " ");
    System.out.println();
    */
    ArrayList<Integer> ret = new ArrayList<Integer>();

    while(s > 0){
      if (s - f[i] >= 0){
        s-=f[i];
        ret.add(f[i]);
      }
      i--;
    }
    if (ret.size() == 1){
      ret.add(0);
    }
    System.out.println(ret.size());
    for(int t:ret)
      System.out.print(t + " ");
    System.out.println();
  }

	public static void main(String args[]){
    new Main().run();
	}
}