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>
using namespace std;
int s, k, cnt, f[200], ans[200];
int main() {
  scanf("%d%d", &s, &k);
  f[1] = 1;
  int cnt = 1;
  for (int i = 2; i <= 200; i++) {
    int idx = i;
    cnt++;
    for (int j = 1; j <= k; j++) {
      idx--;
      if (idx == 0) break;
      f[i] += f[idx];
    }
    if (f[i] > s) break;
  }
  int res = s, num = 0;
  for (int i = cnt - 1; i >= 1; i--) {
    if (res >= f[i]) {
      num++;
      ans[num] = f[i];
      res -= f[i];
    }
    if (res == 0) break;
  }
  if (num == 1) {
    num++;
    ans[num] = 0;
  }
  printf("%d\n", num);
  for (int i = 1; i < num; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[num]);
  return 0;
}