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 n, k, i, j, s;
  cin >> s >> k;
  vector<long long> v(200), v1(200);
  v[0] = 1, v1[0] = 1;
  i = 1;
  for (; v[i - 1] <= s; i++) {
    if (i - k - 1 >= 0)
      v[i] = v1[i - 1] - v1[i - k - 1];
    else
      v[i] = v1[i - 1];
    v1[i] = v1[i - 1] + v[i];
  }
  long long L = 1;
  vector<long long> ans(200);
  while (s) {
    long long x = 0;
    for (; v[x] <= s; x++)
      ;
    ans[L++] = v[x - 1];
    s -= v[x - 1];
  }
  cout << L << endl;
  for (i = 0; i < L; i++) cout << ans[i] << " ";
  cout << endl;
}