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:
//package Round_139;

import java.util.*;

public class b {

	public static void main(String args[]){
		Scanner in = new Scanner(System.in);
		long s = in.nextLong();
		long k = in.nextLong();
		long f[] = new long[10001];
		f[0] = 1;		
		int i = 1;
		for (i = 1; i<1000; i++){
			for (int j = i-1; j >= 0 && j>=i-k; j--)
				f[i] += f[j];
			if (f[i] > (int)(1e9+1)) break;
		}
		String ans = "";
//		for (int j = 0; j<=i; j++)
//			System.out.print(f[j]  +" ");
//			System.out.println("First step is over");
			
		int len = 0;
		boolean t[] = new boolean[i+1];
		
		while (s > 0){
			for (int j = i; j >=0; j--)
				if (s - f[j] >= 0 && !t[j]) {
					s -= f[j];
					ans += f[j] + " ";
					t[j] = true;
					len++;
					break;
				}
		}
		System.out.println(len == 1 ? 2 : len);
		if (len == 1)
			System.out.println(ans + "0");
		else
			System.out.println(ans);
	}

}