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;
const double PI = acos(-1.0);
const long long INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MOD = 1000 * 1000 * 1000 + 7;
const int MAX = 10000 + 47;
long long F[MAX];
vector<int> R;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int s, k;
  cin >> s >> k;
  int p = 0;
  F[0] = 1;
  for (int i = (1); i < (MAX); i++) {
    for (int j = (max(0, i - k)); j < (i); j++) {
      F[i] += F[j];
    }
    if (F[i] > s) {
      p = i;
      break;
    }
  }
  for (int i = (p)-1; i >= (0); i--) {
    if (s >= F[i]) {
      s -= F[i];
      R.push_back(F[i]);
    }
  }
  if ((int)R.size() == 1) R.push_back(0);
  cout << (int)R.size() << endl;
  for (int i = (0); i < ((int)R.size()); i++) {
    cout << R[i] << " ";
  }
  cout << endl;
}