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> v;
long long s, k, i = 0;
long long f[1000 * 1000];
int main() {
  cin >> s >> k;
  if (s == 1) {
    cout << 2 << endl;
    cout << 0 << " " << 1;
    return 0;
  }
  if (s == 2) {
    cout << 2 << endl;
    cout << 0 << " " << 2;
    return 0;
  }
  f[0] = 1;
  while (f[i] < s) {
    i++;
    long long j = i - 1;
    while (j >= i - k && f[j] != 0 && j >= 0) {
      f[i] += f[j];
      j--;
    }
  }
  if (f[i] == s) {
    cout << 2 << endl;
    cout << 0 << " " << f[i];
    return 0;
  }
  i--;
  while (s != 0) {
    if (s >= f[i]) {
      v.push_back(f[i]);
      s -= f[i];
    }
    i--;
  }
  cout << v.size() << endl;
  for (long long qi = 0; qi < v.size() - 1; qi++) cout << v[qi] << " ";
  cout << v[v.size() - 1];
}