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;
long long n, i = 0, j = 0, x, t, w, p, q, l, r, count1 = 0, count2 = 0, k = 0,
             first = 0, second = 0, last, sum = 1e9;
const long long mod = 1000003;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long s, d;
  cin >> s >> d;
  vector<long long> f(1000, 0), v;
  f[1] = 1;
  i = 1;
  while (true) {
    if (f[i] > s) break;
    i++;
    for (long long j = max(1ll, (i - d)); j < i; j++) {
      f[i] += f[j];
    }
  }
  for (long long j = i; j >= 1; j--) {
    if (s >= f[j]) {
      v.push_back(f[j]);
      s -= f[j];
    }
    if (s == 0) break;
  }
  if (v.size() <= 1) v.push_back(0);
  cout << v.size() << endl;
  for (auto &x : v) cout << x << " ";
  cout << endl;
  return 0;
}