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;
vector<long long> ans;
long long s, f[100], sum[100];
int k;
int main() {
  cin >> s >> k;
  int i;
  f[0] = f[1] = 1;
  sum[0] = 1;
  sum[1] = 2;
  for (i = 2; f[i - 1] < s && i <= k; i++)
    f[i] = sum[i - 1], sum[i] = sum[i - 1] + f[i];
  for (; f[i - 1] < s; i++) {
    f[i] = sum[i - 1] - sum[i - 1 - k];
    sum[i] = sum[i - 1] + f[i];
  }
  for (--i; s > 0;) {
    while (f[i] > s) i--;
    ans.push_back(f[i]);
    s -= f[i];
    i--;
  }
  if (ans.size() == 1) ans.push_back(0);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
  cout << endl;
}