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 long long NMAX = 110;
int N, K, S;
vector<long long> V;
long long Fib[NMAX], Sum[NMAX];
int main() {
  scanf("%I64d %i", &S, &K);
  Fib[1] = Sum[1] = N = 1;
  for (int i = 2;; ++i) {
    Fib[i] = Sum[i - 1] - Sum[max(0, i - K - 1)];
    if (Fib[i] > 1LL * S) break;
    N++;
    Sum[i] = Sum[i - 1] + Fib[i];
  }
  for (int i = N; i; --i)
    if (Fib[i] <= S) V.push_back(Fib[i]), S -= Fib[i];
  while (V.size() < 2) V.push_back(0);
  printf("%i\n", V.size());
  for (long long i = 0; i < V.size(); ++i) printf("%i ", V[i]);
}