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() {
  int s, k;
  while (cin >> s >> k) {
    vector<int> res;
    if (k > 31) k = 31;
    int rng = (1 << k) - 1;
    if (s <= rng) {
      for (int i = (0), _n = (k); i < _n; i++)
        if (s & (1 << i)) res.push_back(1 << i);
    } else {
      int fb[100];
      fb[0] = 0, fb[1] = 1;
      int i = 2;
      for (; fb[i - 1] <= s; i++) {
        fb[i] = 0;
        int tk = k, ti = i;
        while (ti-- && tk--) fb[i] += fb[ti];
      }
      for (i -= 2; i > 1; i--) {
        if (s - fb[i] > 0) {
          s -= fb[i];
          res.push_back(fb[i]);
          if (s <= rng) break;
        }
      }
      for (int j = (0), _n = (k); j < _n; j++)
        if (s & (1 << j)) res.push_back(1 << j);
    }
    if (res.size() < 2) res.push_back(0);
    cout << res.size();
    for (int i = (0), _n = (res.size()); i < _n; i++)
      cout << (!i ? "\n" : " ") << res[i];
    cout << endl;
  }
  return 0;
}