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 a[1000];
int main() {
  long long k, s;
  cin >> s >> k;
  memset(a, 0, sizeof(a));
  a[0] = 1;
  int i;
  for (i = 1;; i++) {
    int j;
    if (i > k)
      j = i - k;
    else
      j = 0;
    for (; j < i; j++) {
      a[i] += a[j];
    }
    if (a[i] > s) break;
  }
  vector<int> v;
  for (int j = i; s != 0 && j >= 0; j--) {
    if (a[j] <= s) {
      s -= a[j];
      v.push_back(a[j]);
    }
  }
  if (v.size() == 1) v.push_back(0);
  cout << v.size() << endl;
  for (int j = v.size() - 1; j >= 0; j--) cout << v[j] << " ";
}