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;
int s, k, a[55], x = 1;
set<int> st;
int main() {
  scanf("%d%d", &s, &k);
  a[0] = 0, a[1] = 1;
  int i;
  for (i = 2; i <= k; ++i) {
    if (x > s || x > 1e9) break;
    a[i] += x;
    x += a[i];
  }
  for (;; ++i) {
    if (x > s || x > 1e9) break;
    x = 0;
    for (int j = i - k; j < i; ++j) x += a[j];
    a[i] = x;
  }
  for (int j = i; j >= 0; --j) {
    if (a[j] <= s && st.count(a[j]) == 0) s -= a[j], st.insert(a[j]);
  }
  printf("%d\n", (int)st.size());
  for (set<int>::iterator it = st.begin(); it != st.end(); ++it)
    printf("%d ", *it);
  return 0;
}
