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;
bool myfunction(int i, int j) { return (i < j); }
vector<int> p;
int s, k;
int main() {
  cin >> s >> k;
  int a = 1;
  while (a <= s) {
    p.push_back(a);
    a = 0;
    for (int i = p.size() - 1, j = 0; i >= 0 && j < k; --i, ++j) a += p[i];
  }
  vector<int> ret;
  int ans = 1;
  ret.push_back(0);
  for (int i = p.size() - 1; i >= 0; --i)
    if (p[i] <= s) {
      ret.push_back(p[i]);
      ans++;
      s -= p[i];
    }
  cout << ans << endl;
  for (int i = 0; i < ret.size(); ++i) cout << ret[i] << " ";
  return 0;
}