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;
long long s, k, n;
long long a[10000];
vector<long long> v;
int main() {
  while (cin >> s >> k) {
    n = k;
    a[n - k] = 1;
    while (a[n - k] <= s) {
      n++;
      a[n - k] = 0;
      for (int i = 1; i <= k && n - i >= k; i++) {
        a[n - k] += a[n - i - k];
      }
    }
    v.clear();
    v.push_back(0);
    for (int i = n; i >= k && s > 0; i--) {
      if (a[i - k] <= s) {
        s -= a[i - k];
        v.push_back(a[i - k]);
      }
    }
    cout << v.size() << endl;
    for (int i = 0; i < v.size(); i++) {
      if (i > 0) cout << ' ';
      cout << v[i];
    }
    cout << endl;
  }
  return 0;
}