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