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 s, a[105], ans[105], n = 0;
int k;
int main() {
  cin >> s >> k;
  k = min(k, 30);
  a[0] = 0, a[1] = 1;
  for (int i = 2; i <= 50; ++i)
    for (int j = 1; j <= k; ++j)
      if (i - j >= 0) a[i] += a[i - j];
  for (int i = 50; i >= 0; --i)
    if (s >= a[i]) s -= a[i], ans[++n] = a[i];
  cout << n << endl;
  for (int i = 1; i <= n; ++i)
    if (i < n)
      cout << ans[i] << " ";
    else
      cout << ans[i] << endl;
  return 0;
}