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, i;
long long f[45];
vector<int> sum;
int main() {
  cin >> s >> k;
  f[1] = 1;
  f[2] = 1;
  int subt;
  for (i = 3; i < 45; ++i) {
    if (i - k - 1 > 0)
      subt = f[i - k - 1];
    else
      subt = 0;
    f[i] = f[i - 1] * 2 - subt;
    if (f[i] > s) {
      --i;
      break;
    }
  }
  for (int j = i; j > 1; --j) {
    if (s - f[j] >= 0) {
      sum.push_back(f[j]);
      s -= f[j];
    }
  }
  if (sum.size() == 1)
    cout << 2 << endl << sum[0] << " 0";
  else {
    cout << sum.size() << endl;
    for (int j = 0; j < sum.size(); ++j) cout << sum[j] << " ";
  }
  return 0;
}