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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long s, k;
  cin >> s >> k;
  if (s == 1) {
    cout << 2 << endl;
    for (long long i = 0; i < 1; i++) {
      cout << 0 << " ";
    }
    cout << 1 << endl;
    return 0;
  }
  long long fib[10000];
  fib[0] = 0;
  fib[1] = 1;
  long long temp = 1;
  long long bb = -1;
  for (long long i = 2; i < k; i++) {
    fib[i] = temp;
    temp += fib[i];
    if (temp > s) {
      bb = i;
      break;
    }
  }
  long long j, i, sum;
  if (bb == -1) {
    fib[k] = temp;
    i = 0;
    j = k;
    sum = 2 * temp;
    while (fib[j] <= s and j < 1000) {
      j += 1;
      if (j - i + 1 > k) {
        sum -= fib[i];
        i++;
      }
      fib[j] = sum;
      sum += fib[j];
    }
  } else {
    j = bb + 1;
  }
  sum = fib[j - 1];
  i = j - 2;
  long long ans[10000];
  long long h = 1;
  ans[0] = fib[j - 1];
  while (sum < s) {
    if (sum + fib[i] <= s) {
      sum += fib[i];
      ans[h] = fib[i];
      h++;
      i--;
    } else {
      i--;
    }
    if (sum == s) {
      break;
    }
  }
  if (h == 1) {
    cout << 2 << endl;
    cout << 0 << " " << ans[0] << endl;
    return 0;
  }
  cout << h << endl;
  for (long long k = 0; k < h; k++) {
    cout << ans[k] << " ";
  }
  cout << endl;
  return 0;
}