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 f[1000000];
void answer(int s, int i, int ans) {
  if (i < 0) {
    cout << ans << "\n";
    return;
  }
  while (f[i] > s) i--;
  answer(s - f[i], i - 1, ans + 1);
  cout << f[i] << " ";
}
int main() {
  int s, k;
  cin >> s >> k;
  f[0] = 0;
  f[1] = 1;
  f[2] = 1;
  for (int i = 2; i < k; i++) {
    if (f[i] >= s) {
      answer(s, i, 0);
      return 0;
    }
    f[i + 1] = f[i] * 2;
  }
  int i = k;
  while (true) {
    if (f[i] >= s) {
      answer(s, i, 0);
      return 0;
    }
    f[i + 1] = f[i] * 2 - f[i - k];
    i++;
  }
  return 0;
}