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
import java.util.Scanner;
import java.util.Stack;

public class Prob225B {
	public static void main(String[] Args) {
		Scanner scan = new Scanner(System.in);
		int sum = scan.nextInt();
		int k = scan.nextInt();
		int maxI = 0;

		int[] ar = new int[200];
		ar[0] = 1;
		ar[1] = 1;
		for (int i = 2; i < 200; i++) {
			ar[i] = ar[i - 1] * 2;
			if (i - k - 1 >= 0)
				ar[i] -= ar[i - k - 1];
			
			if (ar[i] <= sum)
				maxI = i;
			else
				break;
		}

		int tempSum = ar[maxI];
		Stack<Integer> s = new Stack<Integer>();
		s.add(0);
		s.add(ar[maxI]);
		while (tempSum < sum) {
			while (ar[maxI] > sum - tempSum)
				maxI--;

			s.add(ar[maxI]);
			tempSum += ar[maxI];
		}

		System.out.println(s.size());

		while (!s.isEmpty())
			System.out.print(s.pop() + " ");
	}
}
