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 const N = 1e5 + 10;
long long f[N];
int n, m, ans[N];
int main() {
  int s, k;
  scanf("%d%d", &s, &k);
  int n;
  f[1] = 1;
  for (n = 2; f[n - 1] < s; n++)
    for (int j = max(1, n - k); j <= n - 1; j++) f[n] += f[j];
  int m = 0;
  for (int i = n - 1; i >= 1 && s; i--)
    if (f[i] <= s) ans[++m] = f[i], s -= f[i];
  if (m < 2) ans[++m] = 0;
  printf("%d\n", m);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}