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;
set<long long> ans;
vector<long long> kbonaci;
void bonaci(long long n) {
  for (long long i = kbonaci.size() - 1; i >= 0; --i) {
    if (kbonaci[i] <= n && !ans.count(kbonaci[i])) {
      ans.insert(kbonaci[i]);
      n -= kbonaci[i];
    }
    if (n == 0) {
      break;
    }
  }
}
int main() {
  long long s, k;
  cin >> s >> k;
  long long num = 1;
  long long anc = 0;
  kbonaci.push_back(1), kbonaci.push_back(1);
  while (num < s) {
    num *= 2;
    long long sz = kbonaci.size();
    if (sz - k > 0) {
      num -= kbonaci[anc];
      kbonaci.push_back(num);
      anc++;
    } else {
      kbonaci.push_back(num);
    }
  }
  bonaci(s);
  if (ans.size() == 1) {
    cout << 2 << "\n" << 0 << " ";
    for (auto x : ans) {
      cout << x << " ";
    }
  } else {
    cout << ans.size() << "\n";
    for (auto x : ans) {
      cout << x << " ";
    }
  }
}