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;
const unsigned int max_size = 100;
int main() {
  unsigned int s, k;
  unsigned int i, imax;
  cin >> s >> k;
  unsigned int* mas = new unsigned int[max_size];
  imax = 0;
  mas[0] = 1;
  mas[1] = 1;
  i = 2;
  do {
    mas[i] = mas[i - 1] * 2 - (i < k + 1 ? 0 : mas[i - k - 1]);
    if (mas[i] >= s) {
      imax = i;
    }
    i++;
  } while (imax == 0);
  vector<unsigned int> vect;
  for (i = imax; i >= 0 && s > 0; i--) {
    if (mas[i] <= s) {
      vect.push_back(mas[i]);
      s -= mas[i];
    }
  }
  if (vect.size() == 1) vect.push_back(0);
  cout << vect.size() << '\n';
  for (i = 0; i < vect.size(); i++) {
    cout << vect[i] << (i == vect.size() - 1 ? '\n' : ' ');
  }
  return 0;
}