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 M = 1e9 + 9;
const long long N = 1e5 + 10;
const long long IM = 1e16;
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << " ";
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void PTN(long long n = 1, long long a = 0, long long b = 0,
                long long c = 0) {
  if (n--) cout << a << " ";
  if (n--) cout << b << " ";
  if (n--) cout << c << " ";
  cout << "\n";
}
inline void op() {}
long long fp(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    long long j = 0;
    if (n % i == 0) return i;
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long i, j, x, y, u, l;
  {
    i = j = x = y = u = l = 0;
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    v.push_back(0);
    v.push_back(1);
    for (i = 2;; i++) {
      l = 0;
      for (j = i - 1; j >= i - k; j--) {
        l += v[j];
        if (j == 0) break;
      }
      if (l > n)
        break;
      else
        v.push_back(l);
    }
    x = v[v.size() - 1];
    u = n - x;
    long long z = 0;
    vector<long long> ans;
    if (k != 1) ans.push_back(0);
    ans.push_back(x);
    while (x < n) {
      j = lower_bound(v.begin(), v.end(), u) - v.begin();
      if (v[j] > u) j--;
      x += v[j];
      u = u - v[j];
      ans.push_back(v[j]);
    }
    cout << (ans.size()) << "\n";
    PV(ans);
  }
  return 0;
}