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:
#choose n maximal k bonnacii nums less than n and subtract them 
#till n is decomposed...
#it will decompose..! 
#coz any number can be represented as sum of 
#fibonacci numbers 


def s():
    [s,k] = list(map(int,input().split()))
    d = [0]
    d.append(1)
    t = 0
    r = []
    while t < s:
        t = sum(d[-min(k,len(d)):])
        d.append(t)
    for i in reversed(d):
        if i <= s:
            s -= i
            r.append(i)
            if i == 0:
                break
    print(len(r))
    print(*r)
s()


#0 1 1 2 3 5