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 main() {
  long long n, k, t = 1, it, s, i, ac, a[60] = {0}, b[60] = {0};
  cin >> s >> k;
  a[1] = 1;
  a[2] = 1;
  ac = 2;
  it = k;
  while (a[ac] < s) {
    ac++;
    for (i = 1; i <= it; i++) {
      a[ac] += a[ac - i];
      if (ac - i == 1) i = it + 1;
    }
  }
  ac++;
  while (s != 0) {
    ac--;
    if (a[ac] <= s) {
      b[t] = a[ac];
      t++;
      s -= a[ac];
    }
  }
  t--;
  if (t > 0) {
    cout << t + 1 << endl;
  }
  for (i = 1; i <= t; i++) cout << b[i] << " ";
  cout << "0";
}