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.*;
//import javax.swing.JOptionPane;

public class Solution {
	
	public static void main(String args[])
	{
		long dp[] = new long[101];
		Scanner in = new Scanner(System.in);
		int s,k;		
		s = in.nextInt();
		k = in.nextInt();
		dp[0]=1;
		int count = 0;
		for(int i =1; ;i++)
		{
			int past = i - k;
//			System.out.print(past + " ");
			if(past<0)past=0;
			for(int j =past; j<=(i-1) ;j++)
			{
				dp[i]+=dp[j];
			}
//			System.out.println(dp[i]);
			if(dp[i]>(long)s)break;
			count++;
		}
		int ara[] = new int[101];
		int nums = 0;
		for(int i =count;i>=0;i--)
		{
			if(s>=dp[i]){
				s -=dp[i];
				ara[nums++]=(int)dp[i];
			}
		}
		ara[nums++]=0;
		System.out.println(nums);
		for(int i=0;i<nums;i++)System.out.print(ara[i]+" ");
	}

}