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
#include <bits/stdc++.h>
using namespace std;
long long a[100007];
int s, k, n;
multiset<int, greater<int> > col;
vector<int> ans;
int main() {
  while (~scanf("%d%d", &s, &k)) {
    col.clear();
    ans.clear();
    memset(a, 0, sizeof(a));
    n = 1;
    a[1] = 1;
    a[0] = 0;
    while (a[n] - a[n - 1] <= s) {
      int l = max(n - k, 0);
      a[n + 1] = a[n] - a[l];
      col.insert(a[n + 1]);
      a[n + 1] += a[n];
      n++;
    }
    multiset<int>::iterator it;
    while (s) {
      it = col.lower_bound(s);
      s -= *it;
      ans.push_back(*it);
      col.erase(it);
    }
    while (k != 1 && ans.size() < 2) ans.push_back(0);
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
    puts("");
  }
}
