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 s, k;
unsigned int pibo[60];
int main() {
  cin >> s >> k;
  int i, j;
  pibo[0] = 0;
  pibo[1] = 1;
  for (i = k + 1; i <= k + 49; i++) {
    if (i - 2 * k + 1 < 0) {
      for (j = 0; j < i - k + 1; j++) {
        pibo[i - k + 1] += pibo[j];
      }
    } else {
      for (j = i - 2 * k + 1; j < i - k + 1; j++) {
        pibo[i - k + 1] += pibo[j];
      }
    }
    if (pibo[i - k] > 1000000000) {
      break;
    }
  }
  int leng = i - k;
  int res[50], num = 0;
  for (i = leng - 1; i >= 0; i--) {
    if (pibo[i] <= s && res[num] != pibo[i]) {
      num++;
      res[num] = pibo[i];
      s -= pibo[i];
    }
  }
  cout << num << endl;
  for (i = 1; i <= num; i++) cout << res[i] << " ";
  return 0;
}