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 arr[1000000] = {0};
int sum(vector<int> v) {
  int ret = 0;
  for (int i = 0; i < v.size(); i++) ret += v[i];
  return ret;
}
int main() {
  int i, j, k, s;
  scanf("%d %d", &s, &k);
  for (i = 1, j = k; arr[i - 1] < s; i++, j++) {
    if (j == k || j - 1 == k) {
      arr[i] = 1;
    } else if (2 * k >= j) {
      arr[i] = 2 * arr[i - 1];
    } else {
      arr[i] = 2 * arr[i - 1] - arr[i - k - 1];
    }
  }
  vector<int> vv;
  while (1) {
    vv.push_back(arr[i--]);
    if (sum(vv) > s) vv.pop_back();
    if (sum(vv) == s) break;
  }
  cout << vv.size() << endl;
  for (int i = 0; i < vv.size(); i++) cout << vv[i] << " ";
  return 0;
}