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() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  long long s;
  int k;
  cin >> s >> k;
  vector<long long> arr;
  arr.push_back(0);
  arr.push_back(1);
  arr.push_back(1);
  while (arr.back() < s) {
    long long acc = 0;
    for (int i = max(0, (int)(arr).size() - k); i < ((int)(arr).size()); ++i) {
      acc += arr[i];
    }
    arr.push_back(acc);
  }
  if (arr.back() > s) arr.pop_back();
  vector<long long> selected;
  while ((int)(arr).size() > 0 && s >= 0) {
    if (arr.back() <= s) {
      selected.push_back(arr.back());
      s -= arr.back();
    }
    arr.pop_back();
  }
  cout << (int)(selected).size() << '\n';
  for (int i = 0; i < ((int)(selected).size()); ++i) cout << selected[i] << ' ';
}