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() {
  int s, k;
  cin >> s >> k;
  vector<long long> f(50);
  f[0] = 0;
  f[1] = 1;
  for (int i = 2; i < 50; i++)
    for (int j = 1; j <= k && j <= i; j++) f[i] += f[i - j];
  vector<int> ans;
  int t = 1;
  while (f[t] < s) t++;
  for (int i = t; i >= 0; i--)
    if (f[i] <= s) {
      ans.push_back(f[i]);
      s -= f[i];
    }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
  return 0;
}