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;
multiset<int, greater<int> > col;
vector<int> ans;
int main() {
  int s, k;
  cin >> s >> k;
  int n = 1;
  long long num[100000];
  memset(num, 0, sizeof num);
  num[0] = 0;
  num[1] = 1;
  while (num[n] - num[n - 1] <= s) {
    int l = max(n - k, 0);
    num[n + 1] = num[n] - num[l];
    col.insert(num[n + 1]);
    num[n + 1] += num[n];
    n++;
  }
  multiset<int>::iterator it;
  while (s) {
    it = col.lower_bound(s);
    s -= *it;
    ans.push_back(*it);
    col.erase(it);
  }
  while (ans.size() < 2) ans.push_back(0);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}