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;
long long f[100000] = {0};
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < 100000; i++) f[i] = 0;
  long long tot = 1;
  f[1] = 1;
  while (true) {
    if (f[tot] > n) {
      break;
    } else {
      tot++;
      long long def = 1;
      for (long long j = max(def, tot - k); j < tot; j++) {
        f[tot] += f[j];
      }
    }
  }
  vector<long long> ans;
  ans.push_back(0);
  long long ans1 = 1;
  for (long long i = tot; i >= 1; i--) {
    if (f[i] <= n) {
      ans1++;
      ans.push_back(f[i]);
      n -= f[i];
      if (n <= 0) {
        break;
      }
    }
  }
  cout << ans1 << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}