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;
  cin >> s >> k;
  vector<int> arr;
  int sum = 1;
  int index = 1;
  arr.push_back(1);
  for (int i = 0; i < k; i++) {
    if (sum > s) break;
    arr.push_back(sum);
    index++;
    sum *= 2;
  }
  sum /= 2;
  int last = 0;
  while (sum != 0) {
    sum = sum * 2 - arr[last];
    if (sum >= s) break;
    last++;
    arr.push_back(sum);
    index++;
  }
  index--;
  int usedcount = 0;
  vector<int> used;
  while (s != 0) {
    if (arr[index] <= s) {
      s -= arr[index];
      used.push_back(arr[index]);
      usedcount++;
    }
    index--;
  }
  if (usedcount == 1) {
    cout << usedcount + 1 << endl;
    cout << "0 ";
  } else {
    cout << usedcount << endl;
  }
  for (int i = 0; i < usedcount; i++) {
    cout << used[i] << " ";
  }
}