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;
int n;
int s, k;
int flag[100100];
int ans[100100];
int p, q;
int main() {
  while (~scanf("%d%d", &s, &k)) {
    p = q = 0;
    flag[0] = 0;
    flag[++p] = 1;
    for (int i = 1; i <= k; i++) {
      p++;
      if (i == 1)
        flag[p] = 1;
      else
        flag[p] = flag[p - 1] * 2;
      if (flag[p] >= s) break;
    }
    if (flag[p] >= s) {
      for (int i = p; i && s; i--) {
        if (flag[i] <= s) {
          ans[++q] = flag[i];
          s -= flag[i];
        }
      }
    } else {
      for (int i = 1;; i++) {
        p++;
        flag[p] = flag[p - 1] * 2 - flag[i];
        if (flag[p] >= s) break;
      }
      for (int i = p; i && s; i--) {
        if (flag[i] <= s) {
          ans[++q] = flag[i];
          s -= flag[i];
        }
      }
    }
    if (q >= 2) {
      printf("%d\n", q);
      for (int i = 1; i < q; i++) printf("%d ", ans[i]);
      printf("%d\n", ans[q]);
    } else {
      printf("%d\n", 2);
      printf("%d %d\n", 0, ans[1]);
    }
  }
  return 0;
}
