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 main() {
  int n, k, f[110] = {0};
  scanf("%d%d", &n, &k);
  f[0] = 0;
  f[1] = 1;
  int i = 1;
  while (i++) {
    for (int j = 1; j <= k && i - j > 0; j++) f[i] += f[i - j];
    if (f[i] > n) break;
  }
  i--;
  int tmp[110] = {0}, b = 0;
  for (int j = i; j >= 0; j--) {
    if (n >= f[j]) {
      if ((n - f[j]) != 0 || b != 0 || j > 1) {
        n -= f[j];
        tmp[b] = f[j];
        b++;
      }
    }
  }
  printf("%d\n", b);
  printf("%d", tmp[b - 1]);
  for (int j = b - 2; j >= 0; j--) printf(" %d", tmp[j]);
  printf("\n");
  return 0;
}