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
#include <bits/stdc++.h>
using namespace std;
double const PI = acos(-1.0);
double const EPS = 1e-9;
long long int acu[1000000];
long long int fib[1000000];
long long res[1000];
int n_res;
int cont;
set<pair<int, int> > mem;
void get_sum(int i, int s, int n) {
  if (n_res != -1) {
    return;
  }
  if (i == -1) {
    return;
  }
  if (s == 0) {
    n_res = n;
    return;
  }
  if (s - fib[i] >= 0) {
    res[n] = fib[i];
    get_sum(i - 1, s - fib[i], n + 1);
  }
  if (n_res == -1) get_sum(i - 1, s, n);
}
int main() {
  int s;
  int k;
  scanf("%d %d", &s, &k);
  k++;
  cont = 0;
  unsigned long long int x = 1;
  while (x <= s) {
    x = cont == 0 ? 1 : acu[cont - 1];
    if (cont >= k) x -= acu[cont - k];
    acu[cont] = (cont == 0 ? 0 : acu[cont - 1]) + x;
    fib[cont] = x;
    cont++;
  }
  n_res = -1;
  mem.clear();
  get_sum(cont - 1, s, 0);
  if (n_res < 2) res[n_res++] = 0;
  cout << n_res << endl;
  for (int i = 0; i < n_res; i++)
    cout << res[i] << (i == n_res - 1 ? '\n' : ' ');
  return 0;
}
