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>
int use[48];
int main(void) {
  int s, k;
  int ans[48];
  while (~scanf("%d %d", &s, &k)) {
    if (k > 47) k = 48;
    use[0] = 0;
    use[1] = 1;
    for (int i = 2; i < 48; i++) {
      use[i] = 0;
      for (int j = 1; j <= k && j <= i; j++) {
        use[i] += use[i - j];
      }
    }
    int temp = 0;
    for (int i = 0; i < 48; i++) {
      if (s - use[i] == 0) {
        printf("%d\n", temp + 2);
        ans[temp] = use[i];
        break;
      }
      if (s > use[i] && s < 2 * use[i]) {
        ans[temp] = use[i];
        temp++;
        s = s - use[i];
        i = 0;
      }
    }
    for (int i = 0; i <= temp; i++) {
      if (i == temp) {
        printf("%d 0\n", ans[i]);
        break;
      }
      printf("%d ", ans[i]);
    }
  }
}