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
#include <bits/stdc++.h>
using namespace std;
long long int factorial(int n) {
  if (n == 0 || n == 1) {
    return 1;
  } else {
    return n * factorial(n - 1);
  }
}
void file_i_o() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  file_i_o();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int s, k;
  cin >> s >> k;
  vector<long long int> fib(100001, 0);
  vector<long long int> sum(100001, 0);
  sum[1] = 1;
  fib[1] = 1;
  long long int siz = 0;
  for (long long int i = 2;; i++) {
    if (i <= k) {
      fib[i] = sum[i - 1];
    } else {
      fib[i] = sum[i - 1] - sum[i - k - 1];
    }
    if (fib[i] > s) {
      siz = i;
      break;
    }
    sum[i] = sum[i - 1] + fib[i];
  }
  sum[siz] = sum[siz - 1] + fib[siz];
  vector<long long int> ans(100001, 0);
  long long int t = 0;
  for (int i = siz - 1; i > 0; i--) {
    if (s - fib[i] >= 0) {
      s -= fib[i];
      ans[t] = fib[i];
      t++;
    }
  }
  if (t == 1) {
    t++;
    ans[t] = 0;
  }
  cout << t << endl;
  for (int i = 0; i <= t - 1; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
