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 main() {
  ios_base::sync_with_stdio(false);
  int s, k;
  cin >> s >> k;
  deque<int> q;
  vector<int> v;
  q.push_back(1);
  v.push_back(1);
  int sum = 1;
  for (int i = k + 1; i < k + 100 && sum <= s; i++) {
    q.push_back(sum);
    v.push_back(sum);
    sum += sum;
    if ((int)q.size() > k) {
      if (i >= 2 * k) sum -= q.front();
      q.pop_front();
    }
  }
  int i = v.size() - 1;
  vector<int> ans;
  while (s > 0) {
    if (s >= v[i]) {
      s -= v[i];
      ans.push_back(v[i]);
    }
    i--;
  }
  if ((int)ans.size() == 1) {
    cout << ans.size() + 1 << endl;
    cout << 0 << ' ';
  } else {
    cout << ans.size() << endl;
  }
  for (i = 0; i < (int)ans.size(); i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}