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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long s;
  long long k;
  cin >> s >> k;
  vector<long long> v;
  v.push_back(1);
  long long sum = 1;
  for (long long i = 1; i < 200; i++) {
    if (i < k) {
      v.push_back(sum);
      sum += v[i];
    } else {
      v.push_back(sum);
      sum += v[i] - v[i - k];
    }
    if (sum > 1000000007) break;
  }
  sort(v.begin(), v.end());
  vector<long long> ans;
  long long cnt = 0;
  while (s != 0) {
    long long i = lower_bound(v.begin(), v.end(), s + 1) - v.begin();
    i--;
    cnt++;
    ans.push_back(v[i]);
    s -= v[i];
  }
  if (cnt == 1) {
    cout << 2 << "\n" << 0 << " " << ans[0];
    return 0;
  }
  cout << cnt << "\n";
  for (auto i : ans) {
    cout << i << " ";
  }
}