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, f[10000], t, c, res, a[10000], a_c;
int main() {
  scanf("%d%d", &s, &k);
  t = 1;
  c = 3;
  f[2] = 1;
  for (int i = 3; i <= k + 1; i++) {
    f[i] = t, t += f[i], res += f[i];
    c++;
    if (res > s) break;
  }
  while (1) {
    if (c - k - 1 > 0) t -= f[c - k - 1];
    if (res > s) break;
    f[c] = t;
    res += f[c];
    t += f[c];
    c++;
  }
  for (int i = c; i >= 3; i--)
    if (res - f[i] >= s)
      res -= f[i];
    else
      a[++a_c] = f[i];
  printf("%d\n", a_c + 1);
  printf("0 ");
  for (int i = 1; i <= a_c; i++) printf("%d ", a[i]);
  return 0;
}