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;
const int MaxN = 1000010;
int S, K, N;
long long f[MaxN], s[MaxN];
vector<int> ans;
int main() {
  scanf("%d%d", &S, &K);
  s[1] = f[1] = 1;
  N = 2;
  while ((f[N] = s[N - 1] - s[max(0, N - K - 1)]) <= S) {
    s[N] = s[N - 1] + f[N];
    if (++N > 200) break;
  }
  ans.push_back(0);
  for (int i = N - 1; i > 1; --i)
    if (S >= f[i]) {
      S -= f[i];
      ans.push_back(int(f[i]));
    }
  printf("%d\n", int(ans.size()));
  for (int i = 0; i < int(ans.size()); ++i) printf("%d ", ans[i]);
  return 0;
}