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;
vector<int> f;
vector<int> ans;
int binary_search(int target) {
  int low = 1, high = f.size() - 1, mid;
  while (low + 1 < high) {
    mid = (low + high) / 2;
    if (f[mid] >= target)
      high = mid - 1;
    else
      low = mid;
  }
  if (f[high] < target) return high;
  return low;
}
int main() {
  int s, k;
  cin >> s >> k;
  f.push_back(0);
  f.push_back(1);
  int sum = 1;
  f.push_back(sum);
  while (sum < s) {
    int tmp = 0;
    if ((int)f.size() - 1 - k > 0) tmp = f[f.size() - 1 - k];
    sum = sum * 2 - tmp;
    f.push_back(sum);
  }
  int num = s;
  while (num > 0) {
    int index = binary_search(num);
    if (f[index + 1] == num) {
      ans.push_back(f[index + 1]);
      num = 0;
    } else {
      ans.push_back(f[index]);
      num -= f[index];
    }
  }
  ans.push_back(0);
  cout << ans.size() << endl;
  cout << ans[0];
  for (int i = 1; i < ans.size(); i++) cout << ' ' << ans[i];
  cout << endl;
  return 0;
}