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 maxn = 1000000000;
long long x[10010], s, k;
long long ans[10010];
bool u[10010];
int cnt, c;
void init() {
  cnt = 0;
  x[0] = 1;
  int i;
  for (i = 1; i <= k; i++) {
    ++cnt;
    x[i] = (1 << (i - 1));
    if (x[cnt] >= s) break;
  }
  if (x[cnt] < s) {
    for (i = cnt + 1;; i++) {
      ++cnt;
      x[i] = 2 * x[i - 1] - x[i - k - 1];
      if (x[i] >= s) break;
    }
  }
}
void dfs(long long sum) {
  if (sum <= 0) return;
  int i;
  for (i = 1;; i++) {
    if (x[i] > sum / 2) {
      ans[++c] = x[i];
      break;
    }
  }
  dfs(sum - x[i]);
}
int main() {
  while (cin >> s >> k) {
    init();
    c = 0;
    dfs(s);
    if (c == 1) {
      cout << "2" << endl;
      cout << "0"
           << " " << ans[1] << endl;
      continue;
    }
    cout << c << endl;
    for (int i = 1; i < c; i++) cout << ans[i] << " ";
    cout << ans[c] << endl;
  }
  return 0;
}