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;
const long long N = 100;
long long mass[N];
long long n;
vector<long long> res;
void f(long long s) {
  if (s == 0) return;
  long long i;
  for (i = 0; i < n && mass[i] < s; ++i) {
  }
  if (mass[i] == s) {
    res.push_back(mass[i]);
  } else {
    res.push_back(mass[i - 1]);
    f(s - mass[i - 1]);
  }
}
int main() {
  long long s, k;
  cin >> s >> k;
  k = min(k, 50LL);
  mass[k] = 1;
  long long i;
  for (i = k + 1; i < N && mass[i - 1] < s; ++i) {
    for (long long j = i - k; j < i; ++j) {
      mass[i] += mass[j];
    }
  }
  n = i;
  f(s);
  res.push_back(0);
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); ++i) {
    cout << res[i] << " ";
  }
  return 0;
}