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 k;
long long s;
vector<long long> t, rr;
int main() {
  cin >> s >> k;
  t.push_back(1);
  int tmp = 1;
  t.push_back(tmp);
  while (tmp < s) {
    tmp = 0;
    int u = t.size() - k;
    for (int i = (max(0, u)); i < t.size(); i++) tmp += t[i];
    t.push_back(tmp);
  }
  int cnt = t.size() - 1;
  while (s > 0) {
    if (t[cnt] > s)
      cnt--;
    else {
      s -= t[cnt];
      rr.push_back(t[cnt]);
    }
  }
  sort(rr.begin(), rr.end());
  cout << rr.size() + 1 << endl;
  for (int i = 0; i < rr.size(); i++) cout << rr[i] << ' ';
  cout << "0 ";
  return 0;
}