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 main() {
  int s, k, sum = 1, p = 2, i;
  vector<int> v, w;
  scanf("%d %d", &s, &k);
  v.push_back(0);
  v.push_back(1);
  for (i = 2;; i++) {
    if (sum > s) break;
    v.push_back(sum);
    sum += sum;
    p++;
    if (p > k) {
      p--;
      sum -= v[i - p];
    }
  }
  while (s) {
    for (i = v.size() - 1; i >= 0; i--) {
      if (v[i] > s) continue;
      s -= v[i];
      w.push_back(v[i]);
      break;
    }
  }
  if (w.size() == 1) w.push_back(0);
  printf("%d\n", w.size());
  for (i = 0; i < w.size(); i++) {
    if (i > 0) putchar(' ');
    printf("%d", w[i]);
  }
  puts("");
  return 0;
}