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 b, c, d, e;
long long f[10000];
long long now[10000];
long long ans[1000];
int num = 0, fuck = 0;
long long s, k;
int dfs(long long i, long long sum) {
  if (fuck == 1) return 0;
  if (sum == s) {
    fuck = 1;
    cout << num + 1 << endl;
    cout << 0 << ' ';
    for (int a = num; a > 1; a--) cout << ans[a] << ' ';
    cout << ans[1] << endl;
    return 0;
  }
  if (sum + f[i] < s) return 0;
  if (sum > s) return 0;
  if (f[i] == 0) return 0;
  num++;
  ans[num] = now[i];
  dfs(i - 1, sum + now[i]);
  num--;
  if (fuck == 1) return 0;
  dfs(i - 1, sum);
}
int search(long long i) {
  if (i < 0) return 0;
  return i;
}
int main() {
  cin >> s >> k;
  f[1] = 1;
  now[1] = 1;
  long long a;
  for (a = 2;; a++) {
    now[a] = f[a - 1] - f[search(a - 1 - k)];
    f[a] = f[a - 1] + now[a];
    if (now[a] > s) break;
  }
  long long temp = a;
  dfs(temp - 1, 0);
}