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 b[50], a[50];
  int t, i, j, s, k;
  memset(b, 0, sizeof(b));
  b[1] = 1;
  t = 0;
  cin >> s >> k;
  for (i = 2; i <= 45; i++) {
    if (i - k >= 1)
      b[i] = 2 * b[i - 1] - b[i - k - 1];
    else
      for (j = i - 1; j >= 1; j--) b[i] += b[j];
  }
  for (i = 45; i >= 1; i--) {
    if (s >= b[i]) {
      s -= b[i];
      a[++t] = b[i];
    }
    if (s == 0) {
      if (t == 1) {
        cout << '2' << endl;
        cout << "0 " << a[1];
      } else {
        cout << t << endl;
        for (j = 1; j <= t; j++) cout << a[j] << ' ';
      }
      cout << endl;
      break;
    }
  }
  return 0;
}
