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 a[100000], t, s, i, len, j, cnt, res[100000], m, n, k, sum;
int main() {
  scanf("%d %d", &s, &k);
  a[0] = 1;
  for (i = 1; a[i - 1] <= s; i++) {
    j = i - 1;
    t = k;
    while (j >= 0 && t--) a[i] += a[j--];
  }
  t = i - 1;
  while (s > 0 && t >= 0) {
    if (a[t] <= s) {
      res[m++] = a[t];
      s -= a[t];
    }
    t--;
  }
  if (m == 1)
    printf("2\n0 ");
  else
    printf("%d\n", m);
  for (i = 0; i < m; i++) printf("%d ", res[i]);
  printf("\n");
  return 0;
}