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
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  int pu;
  int s, k;
  cin >> s >> k;
  if (s == 1) {
    cout << 2 << endl << 0 << ' ' << 1 << endl;
    return 0;
  }
  if (s == 2) {
    cout << 2 << endl << 0 << ' ' << 2 << endl;
    return 0;
  }
  vector<int> bon;
  bon.push_back(1);
  for (int i = 0; i < k && bon[bon.size() - 1] < s; i++) {
    pu = pow(2, i);
    if (pu <= s)
      bon.push_back(pu);
    else
      goto res;
  }
  do {
    pu = accumulate(bon.end() - k, bon.end(), 0);
    if (pu <= s) bon.push_back(pu);
  } while (pu < s);
res:
  vector<int> ans;
  int y = 0;
  for (int i = 0; i < bon.size(); i++) {
    if (s == bon[i]) {
      cout << 2 << endl << 0 << ' ' << s << endl;
      return 0;
    }
  }
  while (s) {
    if (bon[bon.size() - 1 - y] <= s) {
      s -= bon[bon.size() - 1 - y];
      ans.push_back(bon[bon.size() - 1 - y]);
    }
    y++;
  }
  cout << ans.size() << endl << ans[0];
  for (int i = 1; i < ans.size(); i++) cout << ' ' << ans[i];
  cout << endl;
  return EXIT_SUCCESS;
}
