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 K;
long long S;
set<long long> s;
vector<long long> v;
long long a[20000];
vector<long long> ans;
void generate(void) {
  int i, j;
  if (K > 10000) K = 10000;
  a[K - 1] = a[K] = 1;
  for (i = K + 1;; i++) {
    a[i] = 2 * a[i - 1] - a[i - 1 - K];
    if (a[i] > S) break;
  }
  for ((j) = 0; (j) < (int)(i); (j)++) s.insert(a[j]);
  for (__typeof((s).begin()) itr = (s).begin(); itr != (s).end(); itr++)
    v.push_back(*itr);
}
int main(void) {
  int i;
  cin >> S >> K;
  generate();
  while (S > 0) {
    for (i = v.size() - 1;; i--)
      if (v[i] <= S) break;
    S -= v[i];
    ans.push_back(v[i]);
  }
  while (ans.size() < 2) ans.push_back(0);
  int sz = ans.size();
  cout << sz << endl;
  for ((i) = 0; (i) < (int)(sz); (i)++) {
    cout << ans[i];
    if (i == sz - 1)
      cout << endl;
    else
      cout << ' ';
  }
  return 0;
}