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;
  scanf("%d %d", &s, &k);
  vector<long long> fk;
  fk.insert(fk.begin(), 1);
  while (fk[0] <= s) {
    long long val = 0;
    for (int i = 0; i < k && i < fk.size(); ++i) val += fk[i];
    fk.insert(fk.begin(), val);
  }
  vector<int> v;
  int i = 1;
  int res = fk[i];
  v.push_back(fk[i]);
  i++;
  while (res != s) {
    if (res + fk[i] <= s) {
      res += fk[i];
      v.push_back(fk[i]);
    }
    i++;
  }
  if (v.size() == 1) v.insert(v.begin(), 0);
  printf("%d\n", v.size());
  for (int i = 0; i < v.size(); ++i) {
    printf("%d%s", v[i], i < v.size() - 1 ? " " : "\n");
  }
  return 0;
}