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;
void solve() {
  long long s, k;
  cin >> s >> k;
  long long sum = 1;
  vector<long long> v;
  for (long long i = 0; i < k && sum <= s; i++) {
    v.push_back(sum);
    sum += v[i];
  }
  if (sum <= s) {
    v.push_back(v.back() * 2 - 1);
    sum = v.back() * 2 - 1;
  }
  for (long long i = 0; sum <= s; i++) {
    v.push_back(v.back() * 2 - v[i]);
    sum = v.back() * 2 - v[i];
  }
  vector<long long> b;
  for (long long i = v.size() - 1; i >= 0; i--) {
    if (s == 0) break;
    if (v[i] <= s) {
      s -= v[i];
      b.push_back(v[i]);
    }
  }
  if (b.size() < 2) b.push_back(0);
  cout << b.size() << '\n';
  for (auto it : b) cout << it << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  while (test--) solve();
}