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.Scanner;


public class p139B {
	public static void main(String[] args){
		Scanner	in = new Scanner(System.in);
		int s = in.nextInt();
		int k = in.nextInt();
		int i, t = 0;
		long []S = new long [100005];
		long []sum = new long [100005];
		long []v = new long [100005];
		S[0] = 1;
		S[1] = 1;
		sum[0] = 0;
		sum[1] = 1;
		sum[2] = 2;
		for(i = 2; ; ++ i){
			if(i <= k){
				S[i] = S[i-1] + S[i-1];
				sum[i] = sum[i-1] + S[i];
			}
			else{
				S[i] = sum[i-1] - sum[i-k-1];
				sum[i] = sum[i-1] + S[i];
			}
			if(S[i] >= s)
				break ;
		}
		for(; ; -- i){
			if(S[i] <= s){
				s -= S[i];
				v[t ++] = S[i];
			}
			if(s == 0)
				break ;
		}
		System.out.println(t+1);
		System.out.print("0 ");
		for(i = 0; i < t; ++ i){
			System.out.print(v[i] + " ");
		}
	}
}