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
#include <bits/stdc++.h>
using namespace std;
long long fib[5000005];
int solve(long long s, int k) {
  if (s == 1) return 1;
  fib[1] = 1;
  fib[2] = 1;
  long long sum = 2, last = 1;
  int i, j;
  int near;
  for (i = 3; i <= k; ++i) {
    fib[i] = sum;
    sum += fib[i];
    last = fib[i];
    if (fib[i] > s) {
      near = fib[i - 1];
      return near;
    }
    if (fib[i] == s) {
      near = s;
      return near;
    }
  }
  if (last < s) {
    for (i = k + 1;; ++i) {
      fib[i] = 0;
      for (j = 1; j <= k; ++j) {
        fib[i] += fib[i - j];
        if (fib[i] > s) break;
      }
      if (j != k + 1) {
        if (fib[i] == s) {
          near = s;
          return near;
        }
        near = fib[i - 1];
        break;
      }
    }
  }
  return near;
}
int ans[5000005];
int main() {
  int k, i, j, len = 0;
  int s;
  long long near;
  scanf("%d %d", &s, &k);
  ans[len++] = 0;
  while (s != 0) {
    near = solve(s, k);
    ans[len++] = near;
    s -= near;
  }
  printf("%d\n%d", len, ans[0]);
  for (i = 1; i < len; ++i) printf(" %d", ans[i]);
  printf("\n");
  return 0;
}
