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 n, i, j, t = 1, s = 1, k, e, l = 1, l2, l3, a, dp[500001], b, mx = 1,
                   A[500001], m, c, p = 1;
vector<long long> verr;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  dp[1] = 1;
  l = 1;
  s = 1;
  while (dp[l] < n) {
    l++;
    if (l - p > k) {
      s = s - dp[p];
      p++;
    }
    dp[l] = s;
    s = s + dp[l];
  }
  while (n > 0) {
    while (dp[l] > n) l--;
    n = n - dp[l];
    verr.push_back(dp[l]);
  }
  if (verr.size() == 1) verr.push_back(0);
  cout << verr.size() << '\n';
  for (i = 0; i < verr.size(); i++) cout << verr[i] << " ";
}