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;
const int MAX = 100010;
int fib[MAX];
static int s1[MAX];
int main(void) {
  long long int s, k, i, j, m = 0;
  cin >> s >> k;
  fib[0] = 0;
  fib[1] = 1;
  fib[2] = 1;
  for (i = 3; i <= k + 1 && fib[i - 1] <= s; i++) fib[i] = 2 * fib[i - 1];
  for (; fib[i - 1] <= s; i++) fib[i] = 2 * fib[i - 1] - fib[i - k - 1];
  i--;
  while (s >= 1) {
    while (fib[i] > s) i--;
    s1[m++] = fib[i];
    s = s - fib[i];
  }
  if (m == 1) m = m + 1;
  cout << m << '\n';
  for (i = 0; i < m; i++) cout << s1[i] << '\t';
  return 0;
}