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 comment(linker, "/STACK:16777216")
using namespace std;
vector<int> mods = {1000000007, 1000000009, 998244353};
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 200005;
const int mod = mods[0];
void task() {
  long long s, k;
  cin >> s >> k;
  std::vector<long long> v, out;
  deque<long long> dq;
  v.push_back(1);
  dq.push_back(1);
  long long cur_sum = 0;
  while (cur_sum < s) {
    cur_sum += dq.back();
    dq.push_back(cur_sum);
    v.push_back(dq.back());
    if (dq.size() > k) {
      cur_sum -= dq.front();
      dq.pop_front();
    }
  }
  bool f = binary_search((v).begin(), (v).end(), s);
  if (f) {
    cout << 2 << "\n";
    cout << 0 << ' ' << s << "\n";
  } else {
    reverse((v).begin(), (v).end());
    for (auto x : v) {
      if (x <= s) {
        out.push_back(x);
        s -= x;
      }
    }
    sort((out).begin(), (out).end());
    cout << out.size() << "\n";
    for (int x : out) {
      cout << x << ' ';
    }
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int __ = 1; __ <= T; __++) {
    task();
  }
  return 0;
}
