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
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
long long s;
int k;
int main() {
  cin >> s >> k;
  v.push_back(0);
  v.push_back(1);
  while (1) {
    long long cur = 0;
    for (int i = (int)v.size() - 1; i >= max((int)v.size() - k, 0); i--)
      cur += v[i];
    if (cur <= 1000 * 1000 * 1000)
      v.push_back(cur);
    else
      break;
  }
  vector<long long> ans;
  for (int i = (int)v.size() - 1; i >= 0; i--)
    if (v[i] <= s) {
      s -= v[i];
      ans.push_back(v[i]);
    }
  cout << (int)ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
