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 s, k, f[2005], t1;
vector<long long> ans;
int main() {
  scanf("%I64d %I64d", &s, &k);
  f[1] = 1;
  t1 = 1;
  while (f[t1] < s) {
    ++t1;
    for (int j = t1 - 1; j >= 1 && j >= t1 - k; --j) f[t1] += f[j];
  }
  for (int i = t1; i >= 1; --i) {
    if (s >= f[i]) {
      s -= f[i];
      ans.push_back(i);
    }
  }
  if (ans.size() < 2) {
    ans.push_back(0);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size() - 1; ++i) {
    printf("%I64d ", f[ans[i]]);
  }
  printf("%I64d\n", f[ans[ans.size() - 1]]);
  return 0;
}