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 B {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        long s = sc.nextLong();
        int k = sc.nextInt();
        long a[] = new long [100];
        int x = 0;
        if(30 < k ){
            long sum = 0;
            a[2] = 1;
            a[1] = 0;
                for (int i = 3; i < 50; i++) {
//                  System.out.println("hb");
                    sum += a[i-1];
                    a[i] = sum;
                    if (sum >= s){
                        x = i-1;
                        break;
                    }
                }
        } else {
            a[k] = 1;
            for (int i = k+1; i < 1000000; i++) {
                long sum = 0;
                for (int j = i-k; j < i; j++) {
                    sum += a[j];
                }
                a[i] = sum;
                if (sum >= s){
                    x = i;
                    break;
                }
            }
        }
//      System.out.println(x+1);
//      ------------------------------------------
        long ans = 0;
        int cnt = 0;
        long b[]= new long[100000];
        for (int i = x; i >= 0; i--) {
            if ((a[i]+ans) <= s){
                ans += a[i];
                cnt++;
                b[cnt] = a[i];
            }
            if (ans == s){
                if (cnt == 1){
                    System.out.println("2");
                    System.out.println("0 "+b[cnt]);
                    return;
                }
                System.out.println(cnt);
                for (int j = 1; j <= cnt; j++) {
                    System.out.print(b[j]+" ");
                }
                return;
            }
        }
    }

}