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, index, current, i, j, sum, count;
  scanf("%d %d", &s, &k);
  vector<long long int> array;
  array.push_back(0);
  array.push_back(1);
  index = 2;
  while (1) {
    sum = 0;
    i = index - 1;
    while (i >= 0 && (i >= (index - k))) {
      sum += array[i];
      i--;
    }
    array.push_back(sum);
    if (array[index] >= s) break;
    index++;
  }
  int final[index + 5];
  current = 0;
  count = 0;
  for (i = index; i >= 0; i--) {
    if (current + array[i] <= s) {
      current += array[i];
      final[count] = array[i];
      count++;
    }
  }
  printf("%d\n", count);
  for (i = 0; i < count; i++) {
    if (i == count - 1)
      printf("%d\n", final[i]);
    else
      printf("%d ", final[i]);
  }
  return 0;
}