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() {
  int s, k;
  cin >> s >> k;
  vector<int> a;
  a.push_back(1);
  a.push_back(1);
  int i = 1;
  while (a[i] <= s) {
    int s1 = 0;
    for (int j = i; j > i - k && j >= 0; j--) s1 += a[j];
    a.push_back(s1);
    i++;
  }
  i--;
  vector<int> b;
  while (s > 0) {
    if (s >= a[i]) {
      b.push_back(a[i]);
      s -= a[i--];
    } else
      i--;
  }
  if (b.size() == 1) b.push_back(0);
  cout << b.size() << endl;
  for (int i = 0; i < b.size(); i++) cout << b[i] << ' ';
  return 0;
}