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 int mod = 1000003;
long double pi = 3.1415926536;
const int mx = 1000003;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long bp(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long p(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return n / i;
    }
  }
  return 1;
}
int hi[1000000];
int main() {
  int s, k;
  cin >> s >> k;
  hi[0] = 0;
  hi[1] = 1;
  for (int i = 2; hi[i - 1] <= s; i++) {
    for (int j = i - 1; j > 0 && j >= i - k; j--) hi[i] += hi[j];
  }
  int u = 100;
  vector<int> vi;
  while (s) {
    if (hi[u] <= s && hi[u] != 0) s -= hi[u], vi.push_back(hi[u]);
    u--;
  }
  vi.push_back(0);
  cout << vi.size() << '\n';
  for (int i = 0; i < vi.size(); i++) cout << vi[i] << " ";
  return 0;
}