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;
vector<int> a, b;
long long t = 1;
int s, k, i;
int main() {
  scanf("%d %d", &s, &k);
  a.push_back(0), a.push_back(1);
  for (int n = 3; t <= s; n++) {
    a.push_back(t);
    t = 0;
    for (i = max(n - k, 0); i < n; i++) t += a[i];
  }
  a.erase(a.begin() + 1);
  for (i = a.size() - 1; i >= 0; i--)
    if (a[i] <= s) {
      b.push_back(a[i]);
      s -= a[i];
    }
  printf("%d\n", b.size());
  for (i = b.size() - 1; i > 0; i--) printf("%d ", b[i]);
  printf("%d\n", b[i]);
  return 0;
}