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 maxint = -1u >> 1;
const double pi = 3.14159265358979323;
const double eps = 1e-8;
long long s, k;
long long f[100], sum[100];
int main() {
  cin >> s >> k;
  sum[0] = 0;
  sum[1] = sum[2] = 1;
  for (int i = 2; i <= 80; i++) {
    sum[i] = 0;
    for (int j = i, t = 1; j - t >= 1 && t <= k; t++) {
      sum[i] += sum[j - t];
      if (sum[i] > 1000000000) break;
    }
  }
  vector<long long> ans;
  int i;
  while (s) {
    for (i = 1; i <= 80; i++) {
      if (sum[i] > s) break;
    }
    s -= sum[i - 1];
    ans.push_back(sum[i - 1]);
  }
  if (ans.size() == 1) ans.push_back(0);
  cout << ans.size() << endl;
  for (int i = (0); i < (ans.size()); i++) {
    cout << ans[i];
    if (i == ans.size() - 1)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}