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;
map<int, int> ans;
vector<int> ans1;
int main() {
  int s, k;
  cin >> s >> k;
  ans[0] = 1;
  int cur = 1;
  for (int i = 1; cur <= s; i++) {
    int t = 0;
    for (int l = i - 1, tt = 0; l >= 0 && tt < k; tt++, l--) t += ans[l];
    ans[i] = t;
    cur = t;
  }
  for (map<int, int>::reverse_iterator it = ans.rbegin(); it != ans.rend();
       it++) {
    if (it->second && s >= it->second && s) {
      ans1.push_back(it->second);
      s -= it->second;
    }
  }
  ans1.push_back(0);
  cout << ans1.size() << endl;
  sort(ans1.begin(), ans1.end());
  for (int i = 0; i < ans1.size(); i++) cout << ans1[i] << " ";
  return 0;
}