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 F[101], ans[101];
int main() {
  int s, k;
  cin >> s >> k;
  F[0] = 0;
  F[1] = 1;
  F[2] = 1;
  if (k >= 101)
    for (int i = 3; i < 101; i++) {
      F[i] = 2 * F[i - 1];
      if (F[i] > 1000000000) break;
    }
  else {
    for (int i = 3; i <= k; i++) F[i] = 2 * F[i - 1];
    for (int i = k + 1; i < 101; i++) {
      F[i] = 2 * F[i - 1] - F[i - k - 1];
      if (F[i] > 1000000000) break;
    }
  }
  int nn = 0;
  while (s > 0) {
    for (int i = 1; i < 101; i++) {
      if (F[i] > s) {
        ans[nn++] = F[i - 1];
        s -= F[i - 1];
        break;
      }
    }
  }
  cout << nn + 1 << endl;
  cout << "0"
       << " ";
  for (int i = 0; i < nn; i++) cout << ans[i] << " ";
  return 0;
}