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 kbo[1001];
int main() {
  int s, k, i;
  scanf("%d%d", &s, &k);
  kbo[0] = 0;
  kbo[1] = 1;
  kbo[2] = 1;
  for (i = 3;; i++) {
    if (i - k - 1 > 0)
      kbo[i] = 2 * kbo[i - 1] - kbo[i - k - 1];
    else
      kbo[i] = 2 * kbo[i - 1];
    if (kbo[i] > s) break;
  }
  for (int j = 0; j < i; ++j) {
  }
  int res[100], cnt = 0;
  while (s > 0) {
    i--;
    if (s - kbo[i] >= 0) {
      s -= kbo[i];
      res[cnt++] = kbo[i];
    }
  }
  if (cnt == 1)
    printf("2\n0 %d", res[0]);
  else {
    printf("%d\n", cnt);
    for (int j = 0; j < cnt; j++) printf("%d ", res[j]);
  }
  return 0;
}