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;
long long num[1001], sum[1001];
vector<long long> res;
int main() {
  cin >> s >> k;
  num[0] = 1;
  sum[0] = 1;
  int i = 0;
  while (num[i] < s) {
    num[i + 1] = sum[i] - ((i - k >= 0) ? sum[i - k] : 0);
    sum[i + 1] = sum[i] + num[i + 1];
    i++;
  }
  for (i; i >= 0 and s > 0; i--) {
    if (s >= num[i]) {
      s -= num[i];
      res.push_back(num[i]);
    }
  }
  if (res.size() == 1) {
    res.push_back(0);
  }
  cout << res.size() << endl;
  for (i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
  return 0;
}