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;
bool ansflag = false;
vector<long long> fib;
vector<long long> path;
void recur(int n, int sum) {
  if (sum == 0) {
    cout << ((int)((path).size())) + 1 << endl << "0 ";
    for (int i = 0; i < ((int)((path).size())); ++i) {
      cout << path[i] << " ";
    }
    cout << endl;
    ansflag = true;
  } else {
    if (n == 0) return;
    if (sum - fib[n] >= 0) {
      path.push_back(fib[n]);
      recur(n - 1, sum - fib[n]);
      path.pop_back();
    }
    if (ansflag) return;
    recur(n - 1, sum);
  }
}
int main(int argc, char *argv[]) {
  cin >> S >> K;
  fib.push_back(1);
  fib.push_back(1);
  for (int i = 2;; ++i) {
    long long tmp = 0;
    for (int j = i - 1; j >= i - K && j >= 0; --j) {
      tmp += fib[j];
    }
    if (tmp > S) break;
    fib.push_back(tmp);
  }
  recur(((int)((fib).size())) - 1, S);
  return 0;
}