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 zero = 0;
long long one = 1;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long expo(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % 1000000007;
    y = y >> 1;
    x = (1ll * x * x) % 1000000007;
  }
  return res;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
bool prime(long long n) {
  long long i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  if (a.second == b.second) return a.first < b.first;
  return (a.second < b.second);
}
long long rr[] = {
    0, 1, 1, 1, 0, -1, -1, -1,
};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
vector<long long> v;
signed main() {
  long long s, k;
  cin >> s >> k;
  v.push_back(0);
  v.push_back(1);
  while (v.back() < s) {
    long long sum = 0;
    for (long long i = max(0, (long long)v.size() - k); i < v.size(); i++)
      sum += v[i];
    v.push_back(sum);
  }
  vector<long long> ans;
  for (long long i = v.size() - 1; i >= 0; i--)
    if (s >= v[i]) {
      ans.push_back(v[i]);
      s -= v[i];
    }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
}