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 k, s, luu;
int F[100000];
int a[100000];
int res;
int main() {
  cin >> s >> k;
  F[0] = 1;
  F[1] = 1;
  for (int i = 2; i <= 10000; i++) {
    F[i] = 2 * F[i - 1];
    if (i >= k + 1) F[i] -= F[i - k - 1];
    if (F[i] > 1000000000) {
      luu = i;
      break;
    }
  }
  for (int i = luu; i >= 0; i--)
    if (s >= F[i]) {
      res++;
      a[res] = F[i];
      s -= F[i];
    }
  res++;
  cout << res << endl;
  cout << 0 << " ";
  for (int i = 1; i < res; ++i) cout << a[i] << " ";
  return 0;
}