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>
long long a[100] = {1}, ans[100];
int main() {
  int s, k, i = 0, cnt = 0;
  scanf("%d%d", &s, &k);
  long long d = 1, sum = 0;
  while (a[i] < s) {
    if (i < k) {
      a[++i] = d;
      d += a[i];
    } else {
      a[i + 1] = 2 * a[i] - a[i - k];
      ++i;
    }
  }
  if (s == a[i]) {
    printf("2\n%I64d 0\n", a[i]);
    return 0;
  }
  --i;
  while (sum < s) {
    if (sum + a[i] <= s) {
      sum += a[i];
      ans[cnt++] = a[i];
    }
    --i;
  }
  printf("%d\n%I64d", cnt, ans[0]);
  for (i = 1; i < cnt; ++i) printf(" %I64d", ans[i]);
  puts("");
  return 0;
}