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>
long kf[1000];
long ch[1000];
int main() {
  long s, k;
  int m = 1;
  long tmp;
  int i, j;
  int n;
  int alone = 0;
  scanf("%ld%ld", &s, &k);
  kf[1] = 1;
  i = 1;
  while (kf[i] <= s) {
    i++;
    tmp = i - k;
    if (tmp < 0) tmp = 0;
    for (j = i - 1; j >= tmp; j--) kf[i] += kf[j];
  }
  while (s > 0) {
    while (kf[i] > s) i--;
    s = s - kf[i];
    ch[m] = kf[i];
    m++;
    i--;
  }
  if (m == 2) alone++;
  if (alone) {
    printf("%d\n", m);
    for (i = 1; i <= m; i++) printf("%ld ", ch[i]);
  } else {
    printf("%d\n", m - 1);
    for (i = 1; i < m; i++) printf("%ld ", ch[i]);
  }
  printf("\n");
  return 0;
}