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 PYTHON3):

Solution:

s,k = map(int,input().split())


if k>=32:
    k_bon = [0]
    for i in range(33):
        k_bon.append(2**i)
else:
    k_bon = [0]*(k-1)
    k_bon.append(1)
    i = k+1
    c = 0
    while len(k_bon)<=100:
        next = sum(k_bon[c:i])
        k_bon.append(next)
        i+=1
        c+=1

k_bon = set(k_bon)
i = 0
ans = []
k_bon = sorted(list(k_bon))
# print(k_bon)
while s!=0:
    if k_bon[i] >= s:
       if s == k_bon[i]:
        s-=k_bon[i]
        # print(s)
        ans.append(k_bon[i])
        i = 0
       else:

        s-=k_bon[i-1]
        # print(s)
        ans.append(k_bon[i-1])
        i = 0

    else:
        i+=1
if len(ans) == 1:
    ans.append(0)
print(len(ans))
print(*ans)