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() {
  long long r, k, sum = 1, sum2 = 1, i = 0, t;
  cin >> r >> k;
  vector<long long> s;
  vector<long long> out;
  s.push_back(0);
  i = k;
  t = 1;
  s.push_back(1);
  while (8) {
    i++;
    if (i <= 2 * k) {
      s.push_back(sum);
      sum2 += sum;
      sum += sum;
    } else {
      sum -= s[t];
      s.push_back(sum);
      sum2 += sum;
      sum += sum;
      t++;
    }
    if (sum2 > r) break;
  }
  sum = s[s.size() - 1];
  out.push_back(sum);
  for (int j = s.size() - 2; j >= 0; j--) {
    sum += s[j];
    if (sum > r)
      sum -= s[j];
    else {
      out.push_back(s[j]);
      if (sum == r) break;
    }
  }
  cout << out.size() << endl;
  for (int i = 0; i < out.size(); i++) cout << out[i] << " ";
}
