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 s, k;
int f[1000005];
vector<int> vec;
int sum = 1, m = 0, t = 0;
void k_bonacci() {
  f[0] = 1;
  int sum = 1;
  for (int i = 1;; i++) {
    if (sum > s) break;
    f[i] = sum;
    sum += f[i];
    if (i + 1 > k) sum -= f[i - k];
    t++;
  }
}
void solve() {
  int j = t;
  while (s) {
    if (s >= f[j]) {
      s -= f[j];
      vec.push_back(f[j]);
    }
    j--;
  }
  if (vec.size() == 1)
    cout << 2 << endl << 0 << " " << *vec.begin() << endl;
  else {
    cout << vec.size() << endl;
    for (vector<int>::iterator i = vec.begin(); i != vec.end(); i++)
      cout << *i << " ";
    cout << endl;
  }
}
int main() {
  cin >> s >> k;
  k_bonacci();
  solve();
  return 0;
}