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>
const int MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
using namespace std;
long long s, k, tab[10007];
long long wynik;
int main() {
  cin >> s >> k;
  if (s == 1) {
    cout << "2" << endl << "1 0" << endl;
    return 0;
  }
  int i;
  list<int> l;
  tab[0] = 1;
  for (i = 1; tab[i - 1] < s; i++) {
    for (int j = max(i - k, 0LL); j < i; j++) tab[i] += tab[j];
  }
  for (int j = i; s > 0 && j >= 0; j--)
    if (tab[j] > 0 && s - tab[j] >= 0) {
      s -= tab[j];
      l.push_back(tab[j]);
    }
  l.push_back(0);
  printf("%d\n", l.size());
  for (typeof(l.begin()) it = l.begin(); it != l.end(); ++it)
    printf("%d ", *it);
  printf("\n");
  return 0;
}