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;
map<int, unsigned long long> fibs;
int k;
unsigned long long f(int n) {
  if (fibs[n] != 0) {
    return fibs[n];
  }
  if (n < k) {
    return 0;
  }
  if (n == k) {
    fibs[n] = 1;
    return 1;
  }
  unsigned long long sum = 0;
  for (int i = 1; i <= k; ++i) {
    if (n - i < k) break;
    sum += f(n - i);
  }
  fibs[n] = sum;
  return sum;
}
int main() {
  int s;
  cin >> s >> k;
  int nums = 0;
  vector<unsigned long long> fs;
  for (int i = k; s != 0; ++i) {
    unsigned long long n = f(i);
    if (n == s) {
      s -= n;
      fs.push_back(n);
      nums++;
      break;
    }
    if (n > s) {
      unsigned long long h = f(i - 1);
      s -= h;
      nums++;
      fs.push_back(h);
      i = k;
    }
  }
  if (nums == 1) {
    fs.push_back(0);
    nums++;
  }
  cout << nums << endl;
  for (int i = 0; i < fs.size(); ++i) {
    cout << fs[i];
    if (i != fs.size() - 1) {
      cout << ' ';
    }
  }
  cout << endl;
  return 0;
}