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 main() {
  long long s, k;
  while (cin >> s >> k) {
    vector<long long> sol;
    if (k >= 32) {
      for (int i = 0; i <= 32; ++i) {
        if (s & (1LL << i)) {
          sol.push_back(1LL << i);
        }
      }
    } else {
      vector<long long> F(k, 0);
      F[k - 1] = 1;
      while (F.back() < s) {
        long long f = 0;
        for (int i = 1; i <= k; ++i) {
          f += F[(int)F.size() - i];
        }
        F.push_back(f);
      }
      for (int i = (int)F.size() - 1; i >= 0; --i) {
        if (F[i] == 0) break;
        if (s >= F[i]) {
          s -= F[i];
          sol.push_back(F[i]);
        }
      }
    }
    if (sol.size() == 1) {
      sol.push_back(0);
    }
    cout << sol.size() << endl;
    for (int i = 0; i < (int)sol.size(); ++i) {
      cout << sol[i];
      if (i + 1 == (int)sol.size()) {
        cout << endl;
      } else {
        cout << ' ';
      }
    }
  }
  return 0;
}