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;
int main() {
  long long s, k;
  cin >> s >> k;
  long long fib[51] = {0};
  fib[0] = 0;
  fib[1] = 1;
  for (int i = 2; i < 51; i++) {
    for (int j = i - 1; j >= 0 && j >= i - k; j--) {
      fib[i] += fib[j];
    }
  }
  vector<long long> r;
  bool tk[51] = {false};
  while (s > 0) {
    for (int i = 50; i >= 0; i--) {
      if (!tk[i] && fib[i] <= s) {
        s -= fib[i];
        r.push_back(fib[i]);
        tk[i] = true;
      }
    }
  }
  if (r.size() == 1) r.push_back(0);
  cout << r.size() << endl;
  for (int i = 0; i < r.size(); i++) {
    cout << r[i];
    if (i + 1 < r.size()) cout << " ";
  }
  cout << endl;
}