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>
#pragma comment(linker, "/STACK:300000000")
#pragma warning(disable : 4800)
using namespace std;
void showTime() { cerr << (double)clock() / CLOCKS_PER_SEC << endl; }
const double pi = 3.1415926535897932384626433832795;
template <class T>
T abs(const T &a) {
  return a >= 0 ? a : -a;
};
template <class T>
T sqr(const T &x) {
  return x * x;
}
int s, k;
int main() {
  cin >> s >> k;
  vector<long long> v;
  v.push_back(1);
  while (true) {
    long long x = 0;
    for (int i = max(0, int((v).size()) - k); i < int((v).size()); i++)
      x += v[i];
    if (x > s) break;
    v.push_back(x);
  }
  long long sum = 0;
  vector<long long> ans;
  for (int i = int((v).size()) - 1; i >= 0; i--) {
    if (sum + v[i] <= s) {
      sum += v[i];
      ans.push_back(v[i]);
    }
    if (sum == s) break;
  }
  if (int((ans).size()) == 1) ans.push_back(0);
  cout << int((ans).size()) << endl;
  for (int i = 0; i < (int((ans).size())); i++) cout << ans[i] << ' ';
}