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 f[1000];
int ans[1000];
int main() {
  int s, k;
  scanf("%d %d", &s, &k);
  f[0] = 1;
  int total = 1;
  for (int i = 1;; ++i) {
    bool ok = false;
    long long t = 0;
    for (int z = k, j = i - 1; z > 0 && j >= 0; --j, --z) {
      t += f[j];
      if (t > s || t < 0) {
        ok = true;
        break;
      }
    }
    if (ok) break;
    f[i] = t;
    ++total;
  }
  int cnt = 0;
  for (int j = total - 1; s > 0; --j) {
    if (f[j] <= s) {
      s -= f[j];
      ans[cnt++] = f[j];
    }
  }
  ans[cnt++] = 0;
  printf("%d\n", cnt);
  for (int i = cnt - 1; i >= 0; --i) printf("%d ", ans[i]);
  return 0;
}