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
#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 1, 0, 0, 0, 0, 0, 0};
int dy[] = {0, 0, 1, -1, 0, 0, 0, 0};
const long long inf = 100000007;
const double eps = 0.000000009;
long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long s, k;
  cin >> s >> k;
  vector<long long> a;
  a.push_back(0);
  a.push_back(1);
  long long sum = 1;
  long long tt = 1e9 + 5;
  for (int i = 2; i <= k; i++) {
    a.push_back(pow(2, i - 2));
    if (pow(2, i - 2) > tt) break;
  }
  while (a[a.size() - 1] < tt) {
    long long x = 0;
    for (int i = 0; i < k; i++) {
      x += a[a.size() - 1 - i];
    }
    a.push_back(x);
  }
  vector<long long>::iterator it;
  vector<long long> sol;
  while (s) {
    it = lower_bound(a.begin(), a.end(), s);
    if (*it > s) it--;
    s -= *it;
    sol.push_back(*it);
  }
  if (sol.size() == 1) {
    cout << 2 << endl << 0 << " " << sol[0] << endl;
  } else {
    cout << sol.size() << endl;
    for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
    cout << endl;
  }
  return 0;
}
