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 CPP):

Solution:
#include <bits/stdc++.h>
int s, k;
long long f[1000], res[1000];
int main() {
  scanf("%d %d", &s, &k);
  f[0] = 0;
  f[1] = 1;
  int i;
  for (i = 2; f[i - 1] <= s; i++)
    for (int j = 1; i - j > 0 && j <= k; j++) f[i] += f[i - j];
  int cnt = 0;
  for (i = i - 1; i >= 0; i--)
    if (f[i] <= s) {
      s -= f[i];
      res[cnt++] = f[i];
    }
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%d ", res[i]);
  return 0;
}