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;
long long s, k, t = 2, counts;
long long out[1000];
vector<int> v;
int main() {
  cin >> s >> k;
  out[1] = 1;
  if (k <= 32) {
    while (out[t - 1] <= s) {
      for (int i = 1; i <= k; i++) {
        if ((t - i) >= 0) {
          out[t] += out[t - i];
        }
      }
      t++;
    }
  } else {
    while (out[t - 1] <= s) {
      out[t] = (long long)pow((double)2, (double)(t - 2));
      t++;
    }
  }
  for (int i = t - 1; i >= 0; i--) {
    if (s >= out[i]) {
      s = s - out[i];
      v.push_back(out[i]);
      counts++;
    }
  }
  cout << counts << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  return 0;
}