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 OO = 1e9;
const long double EPS = (1e-12);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  for (;;) {
    if (a == 0) return b;
    b %= a;
    if (b == 0) return a;
    a %= b;
  }
}
unsigned long long lcm(unsigned long long a, unsigned long long b) {
  unsigned long long temp = gcd(a, b);
  return temp ? (a / temp * b) : 0;
}
pair<long long, pair<long long, long long> > extendedEuclid(long long a,
                                                            long long b) {
  if (a == 0) return make_pair(b, make_pair(0, 1));
  pair<long long, pair<long long, long long> > p;
  p = extendedEuclid(b % a, a);
  return make_pair(
      p.first,
      make_pair(p.second.second - p.second.first * (b / a), p.second.first));
}
long long modInverse(long long a, long long m) {
  return (extendedEuclid(a, m).second.first + m) % m;
}
long long c(long long n, long long k, long long p) {
  long long numerator = 1;
  int i;
  for (i = 0; i < k; i++) numerator = (numerator * (n - i)) % p;
  long long denominator = 1;
  for (i = 1; i <= k; i++) denominator = (denominator * i) % p;
  long long res = modInverse(denominator, p);
  res = (res * numerator) % p;
  return res;
}
vector<long long> vv;
void runEratosthenesSieve(long long upperBound) {
  long long upperBoundSquareRoot = (long long)sqrt((double)upperBound);
  bool *isComposite = new bool[upperBound + 1];
  memset(isComposite, 0, sizeof(bool) * (upperBound + 1));
  for (long long m = 2; m <= upperBoundSquareRoot; m++) {
    if (!isComposite[m]) {
      vv.push_back(m);
      for (long long k = m * m; k <= upperBound; k += m) isComposite[k] = true;
    }
  }
  for (long long m = upperBoundSquareRoot; m <= upperBound; m++)
    if (!isComposite[m]) vv.push_back(m);
}
long long mypow(long long x, long long n) {
  long long y = 1;
  for (int i = 0; i < (int)(n); ++i) y *= x;
  return y;
}
int fib(int k, int n) {
  if (n < k) return 0;
  if (n == k) return 1;
  int sum = 0;
  for (int i = (n - 1); i >= (int)(n - k); --i) sum += fib(k, i);
  return sum;
}
int main() {
  ios::sync_with_stdio(false);
  long long k, s;
  cin >> s >> k;
  vector<long long> v(100);
  v[0] = 0;
  v[1] = 1;
  v[2] = 1;
  int i = 3;
  while (v[i - 1] < s) {
    if (i < k + 1)
      v[i] = v[i - 1] * 2;
    else
      v[i] = v[i - 1] * 2 - v[i - k - 1];
    i++;
  }
  vector<int> ans;
  for (int j = (i - 1); j >= (int)(0); --j) {
    if (v[j] <= s) {
      s -= v[j];
      ans.push_back(v[j]);
    }
  }
  cout << ans.size() << endl;
  for (int j = 0; j < ((int)((ans).size())); ++j) cout << ans[j] << " ";
  return 0;
}