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 f[10001], s, k;
int main() {
  memset(f, 0, sizeof(f));
  cin >> s >> k;
  f[1] = 1;
  int vt;
  for (int i = 2; i <= 1000; i += 1) {
    vt = i;
    f[i] = 0;
    for (int j = 1; j <= k; j += 1) {
      f[i] += f[i - j];
      if ((i - j) == 1) break;
    }
    if (f[i] > s) break;
  }
  if (f[vt] > s) --vt;
  if (f[vt] == s) {
    cout << 2 << endl << 0 << " " << s << endl;
  } else {
    vector<int> th;
    while (1) {
      s -= f[vt];
      th.push_back(f[vt]);
      int x = vt;
      for (int i = 1; i <= x; i += 1)
        if (f[i] > s) {
          vt = i;
          break;
        }
      --vt;
      if (s == 0) {
        cout << th.size() << endl;
        for (int i = 0; i <= th.size() - 2; i += 1) cout << th[i] << " ";
        cout << th[th.size() - 1] << endl;
        break;
      }
      if (vt == 0) break;
    }
  }
  return 0;
}
