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 K, S;
int Push;
int Kar[400];
long long Sum[400];
int St[400];
int main() {
  scanf("%d%d", &S, &K);
  int min = K - 1;
  Kar[1] = 1;
  Sum[1] = 1;
  int sin;
  long long tmp;
  for (sin = K + 1; sin; sin++) {
    tmp = Sum[sin - 1 - min] - Sum[max(0, sin - K - 1 - min)];
    if (tmp > S) break;
    Kar[sin - min] = tmp;
    Sum[sin - min] = Kar[sin - min] + Sum[sin - 1 - min];
  }
  for (int i = sin - 1, s = S; i > 0 && s; i--)
    if (Kar[i - min] <= s) {
      s -= Kar[i - min];
      St[++Push] = Kar[i - min];
    }
  if (Push == 1) St[++Push] = 0;
  printf("%d\n", Push);
  for (int i = Push; i; i--) printf("%d ", St[i]);
  printf("\n");
  return 0;
}