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 s, k;
long long f[10005];
int main(int ac, char **av) {
  cin >> s >> k;
  f[0] = 1;
  int t;
  for (int i = 1;; ++i) {
    for (int j = 1; j <= k; ++j) {
      if (i < j) break;
      f[i] += f[i - j];
    }
    t = i;
    if (f[i] >= s) break;
  }
  vector<int> res;
  for (; t >= 0; --t)
    if (s >= f[t]) {
      res.push_back(f[t]);
      s -= f[t];
      if (!s) break;
    }
  if (res.size() < 2) res.push_back(0);
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); ++i) cout << res[i] << " ";
  return 0;
}