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 s, k;
  scanf("%d%d", &s, &k);
  long long a[110] = {};
  a[0] = 1;
  for (int i = 1; a[i - 1] < s; i++) {
    int sum = 0;
    for (int j = max(0, i - k); j < i; j++) sum += a[j];
    a[i] = sum;
  }
  vector<long long> v;
  while (s > 0) {
    int ii = 0;
    for (int i = 0; a[i] > 0 && a[i] <= s; i++) ii = i;
    s -= a[ii];
    v.push_back(a[ii]);
  }
  if (v.size() <= 1) v.push_back(0);
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
}