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 s, k;
long long a[5000];
long long ans[5000];
int t;
int dfs(int n, int m) {
  if (n == 0) {
    if (t == 1) {
      cout << 2 << endl;
      cout << 0 << ' ' << ans[0] << endl;
    } else {
      cout << t << endl;
      for (int i = t - 1; i >= 0; i--) cout << ans[i] << ' ';
      cout << endl;
    }
    exit(0);
  }
  for (int i = m; i >= 0; i--) {
    if (n >= a[i]) {
      ans[t++] = a[i];
      dfs(n - a[i], i - 1);
      t--;
    }
  }
}
int main() {
  cin >> s >> k;
  int m = 0;
  a[0] = 1;
  for (int i = 1; a[i - 1] < s; i++) {
    for (int j = i - 1; j >= 0 && j >= i - k; j--) a[i] += a[j];
    m = i;
  }
  if (a[m] > s) m--;
  dfs(s, m);
}