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;
using ll = long long;
template <typename T>
using vc = vector<T>;
ll s, k, A[(ll)1e5 + 10], t, tp;
int main() {
  scanf("%lld %lld", &s, &k);
  tp = k - 1;
  while (t < s) {
    t = (tp == k - 1 || tp == k)
            ? 1
            : (t + A[tp - k] - (tp - 2 * k >= 0 ? A[tp - 2 * k] : 0));
    A[(tp++) - k + 1] = t;
  }
  set<ll> S{0};
  for (ll i = tp - k + 1; i >= 0; i--) {
    if (s >= A[i]) {
      s -= A[i];
      S.insert(A[i]);
    }
    if (0 == s) {
      break;
    }
  }
  printf("%lld\n", (ll)S.size());
  for (auto &i : S) {
    printf("%lld ", i);
  }
  printf("\n");
  return 0;
}