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 s, k;
int pd[110];
int mark[110];
int main(void) {
  scanf("%d %d", &s, &k);
  int maior;
  pd[k - 1 - (k - 1)] = 0;
  pd[k - (k - 1)] = 1;
  pd[k + 1 - (k - 1)] = 1;
  for (int i = k + 2;; i++) {
    if (i - 1 - k - (k - 1) >= 0)
      pd[i - (k - 1)] = 2 * pd[i - 1 - (k - 1)] - pd[i - 1 - k - (k - 1)];
    else
      pd[i - (k - 1)] = 2 * pd[i - 1 - (k - 1)];
    if (pd[i - (k - 1)] >= s) {
      maior = i;
      break;
    }
  }
  int cont = 0;
  vector<int> vetor;
  while (s) {
    for (int i = maior; i >= k - 1; i--) {
      if (!s) break;
      if (cont == 0) {
        if (s < pd[i - (k - 1)]) {
          maior = i - 1;
          continue;
        }
        if (mark[i - (k - 1)]) continue;
        vetor.push_back(pd[i - (k - 1)]);
        cont++;
        mark[i - (k - 1)] = 1;
        s -= pd[i - (k - 1)];
      } else {
        if (s < pd[i - (k - 1)]) {
          maior = i - 1;
          continue;
        }
        if (mark[i - (k - 1)]) continue;
        vetor.push_back(pd[i - (k - 1)]);
        mark[i - (k - 1)] = 1;
        s -= pd[i - (k - 1)];
        cont++;
      }
    }
  }
  while (cont < 2) {
    cont++;
    vetor.push_back(0);
  }
  printf("%d\n", cont);
  for (int i = 0; i < vetor.size(); i++) printf("%d ", vetor[i]);
  return 0;
}
