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;
long long f[100050];
long long n, k;
long long ans[5000];
int tot;
int main() {
  scanf("%I64d%I64d", &n, &k);
  long long temp = 1;
  f[0] = 1;
  int i;
  for (i = 1;; ++i) {
    f[i] = temp;
    temp += f[i];
    if (i - k >= 0) {
      temp -= f[i - k];
    }
    if (f[i] >= n) break;
  }
  tot = 0;
  for (; i >= 0; --i) {
    if (f[i] > n) continue;
    n -= f[i];
    ans[tot++] = f[i];
    if (n == 0) break;
  }
  if (tot == 1) {
    ans[tot++] = 0;
  }
  printf("%d\n", tot);
  for (int i = 0; i < tot; ++i) {
    if (i) printf(" ");
    printf("%I64d", ans[i]);
  }
  puts("");
  return 0;
}