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;
vector<long long> vt, ans;
int main() {
  int s, k;
  long long t = 1;
  scanf("%d%d", &s, &k);
  vt.push_back(1);
  for (int i = 1; i < k && t <= s; i++, t <<= 1) vt.push_back(t);
  while (true) {
    t = 0;
    for (int j = 0; j < k && j < vt.size(); j++) t += vt[vt.size() - j - 1];
    if (t > s) break;
    vt.push_back(t);
  }
  for (int i = vt.size() - 1; i >= 0; i--)
    if (vt[i] <= s) {
      s -= vt[i];
      ans.push_back(vt[i]);
    }
  if (ans.size() > 1) {
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) printf("%lld ", ans[i]);
  } else
    printf("2\n%lld 0\n", ans[0]);
  return 0;
}