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 f[(int)(1e6) + 5];
int a[(int)(1e6) + 5];
int i;
int j;
int k;
int cnt = 0;
int s;
int main() {
  cin >> s >> k;
  f[0] = 1;
  for (i = 1; f[i - 1] < s; i++) {
    for (j = i - 1; j >= i - k && j >= 0; j--) {
      f[i] += f[j];
    }
  }
  for (i = (int)(1e6) + 5 - 1; i >= 0 && s != 0; i--) {
    if (s == 0) break;
    if (s - f[i] >= 0 && f[i] > 0) {
      s -= f[i];
      cnt++;
      a[cnt] = f[i];
    }
  }
  if (cnt == 1) {
    cnt++;
    a[cnt] = 0;
  }
  cout << cnt << endl;
  for (i = 1; i <= cnt; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}