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 a[100];
long long sum[100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, n, k, j = 1;
  cin >> n >> k;
  a[1] = sum[1] = 1;
  for (i = 2; i <= 50; i++) {
    if (sum[i - 1] - sum[max(i - k - 1, 0)] > n) break;
    j++;
    a[i] = sum[i - 1] - sum[max(i - k - 1, 0)];
    sum[i] = sum[i - 1] + a[i];
  }
  vector<int> ans;
  for (i = j; i >= 0; i--) {
    if (n >= a[i]) {
      n -= a[i];
      ans.push_back(a[i]);
    }
    if (!a[i]) break;
  }
  cout << ans.size() << "\n";
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}