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;
int main() {
  int M, K;
  scanf("%d%d", &M, &K);
  static int F[1000];
  F[0] = 0;
  F[1] = 1;
  int N = 2;
  while (true) {
    long long Sum = 0;
    for (int i = max(N - K, 0); i < N; i++) Sum += F[i];
    if (Sum > M) break;
    F[N++] = Sum;
  }
  vector<int> Ans;
  int Last = -1;
  while (N--, M)
    if (M >= F[N] && F[N] != Last) {
      Ans.push_back(F[N]);
      M -= Last = F[N];
    }
  if (Ans.size() == 1) Ans.push_back(0);
  printf("%d\n", Ans.size());
  for (int i = 0; i < Ans.size(); i++) {
    printf("%d", Ans[i]);
    putchar((i + 1 == Ans.size()) ? '\n' : ' ');
  }
  return 0;
}