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;
int i, j, k, m, n, l;
long long f[10000 + 10], s[10000 + 10];
bool v[10000 + 10];
vector<int> a, ans;
map<int, int> mp;
bool dfs(int dep, long long val) {
  if (val > n || val + s[((int)(a).size()) - 1] - s[dep - 1] < n) return false;
  if (dep == ((int)(a).size())) {
    if (val == n) {
      for (int i = 0; i < (((int)(a).size())); ++i)
        if (v[i]) ans.push_back(a[i]);
      return true;
    }
    return false;
  }
  v[dep] = true;
  if (dfs(dep + 1, val + a[dep])) return true;
  v[dep] = false;
  if (dfs(dep + 1, val)) return true;
  return false;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    a.clear();
    mp.clear();
    a.push_back(1);
    mp[f[1] = 1]++;
    for (i = 2; f[i - 1] < n; i++) {
      f[i] = 0;
      for (j = i - 1; j > 0 && i - j <= m; --j) f[i] += f[j];
      if (mp[f[i]] == 0) {
        a.push_back(f[i]);
        mp[f[i]]++;
      }
    }
    ans.clear();
    sort(a.begin(), a.end());
    reverse(a.begin(), a.end());
    for (int i = 0; i < (((int)(a).size())); ++i)
      if (a[i] <= n) {
        ans.push_back(a[i]);
        n -= a[i];
      }
    if (((int)(ans).size()) == 1) ans.push_back(0);
    cout << ((int)(ans).size()) << endl;
    for (int i = 0; i < (((int)(ans).size())); ++i)
      if (!i)
        cout << ans[i];
      else
        cout << ' ' << ans[i];
    cout << endl;
  }
  return 0;
}
