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.*;
import java.io.*;
import java.math.*;

public class Main {
	public static Scanner scan = new Scanner(System.in);
	public static boolean bg = false;
	public static HashMap<Long,Long> m1 = new HashMap();
	public static long k = 0;
	public static void main(String[] args) throws Exception {
		int s = Integer.parseInt(scan.next());
		k = Integer.parseInt(scan.next());
		long sum = 0;
		TreeSet<Long> s1 = new TreeSet();
		s1.add(0l);
		for (long i=k+1;;i++){
			sum-=get(i-k-1);
			sum+=get(i-1);
			m1.put(i, sum);
			s1.add(sum);
			if (sum>=s) break;
		}
		if (bg) System.out.println(m1);
		if (bg) System.out.println(s1);
		long cur = s;
		ArrayList<Long> fin = new ArrayList();
		for (;;){
			long f1 = 0;
			f1 = s1.floor(cur);
			fin.add(f1);
			cur-=f1;
			if (f1<=0){
				break;
			}
		}
		System.out.println(fin.size());
		for (long e: fin){
			System.out.print(e+" ");
		}
		System.out.println();
	}
	public static long get (long n1){
		if (n1==0){
			return 0;
		}
		if (n1>=1&&n1<k){
			return 0;
		}
		else if (n1==k){
			return 1;
		}
		else {
			return m1.get(n1);
		}
	}
}
