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:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF225B extends PrintWriter {
	CF225B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF225B o = new CF225B(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] qu = new int[30]; int cnt = 0;
		if (k <= 30) {
			int[] ff = new int[64];
			ff[k] = 1;
			for (int h = k + 1; h < 64; h++) {
				int f = 0;
				for (int h_ = h - k; h_ < h; h_++)
					f = Math.min(f + ff[h_], 1000000001);
				ff[h] = f;
			}
			for (int h = 64 - 1; h >= k; h--)
				if (n >= ff[h]) {
					n -= ff[h];
					qu[cnt++] = ff[h];
				}
		} else {
			for (int h = 0; h < 30; h++)
				if ((n & 1 << h) != 0)
					qu[cnt++] = 1 << h;
		}
		if (cnt == 1)
			cnt++;
		println(cnt);
		while (cnt-- > 0)
			print(qu[cnt] + " ");
		println();
	}
}