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
#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 1e5 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int INFi = 0x7f7f7f7f;
long long test = 1, second, k;
vector<long long> v, ans;
void init() {
  v.push_back(1);
  long long i = 0, j = 0, res = 1;
  while (res <= second) {
    v.push_back(res);
    j += 1;
    res += v[j];
    if (j >= k) {
      res -= v[i];
      i += 1;
    }
  }
}
void solve() {
  cin >> second >> k;
  init();
  while (second > 0) {
    auto it = upper_bound(v.begin(), v.end(), second);
    it--;
    ans.push_back(*it);
    second -= *it;
  }
  ans.push_back(0);
  cout << ans.size() << " "
       << "\n";
  for (auto it : ans) cout << it << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}
