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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m, k = 0;
long long x[100];
vector<int> y;
int main() {
  scanf("%d%d", &n, &m);
  x[1] = 1;
  for (int i = 2;; i++) {
    for (int j = 1; j <= m && i - j >= 0; j++) x[i] += x[i - j];
    if (x[i] > n) break;
    k = i;
  }
  for (int i = k; i >= 1 && x[i] > 0; i--)
    if (x[i] <= n) {
      y.push_back(i);
      n -= x[i];
    }
  if (((int)(y).size()) == 1) y.push_back(0);
  reverse((y).begin(), (y).end());
  printf("%d\n", ((int)(y).size()));
  for (int i = 0; i < ((int)(y).size()); i++) printf("%I64d ", x[y[i]]);
  return 0;
}