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
s, k = map(int, raw_input().split())
F = [1]
total = 1
while len(F) < k and F[-1] <= s :
    val = F[-1]
    F.append(total + val)
    total += val

while F[-1] <= s:
    val = sum(F[-k:])
    F.append(val)
    total += val

C = []
F = F[::-1]
def find_recur(n):
    if n > 0:
        if n in F:
            C.append(n)
        else:
            x = next(x for x in F if x < n)
            if x is not None:
                C.append(x)
                y = find_recur(n - x)
                if y is not None:
                    C.append(y)
    return None
if s in F:
    print 2
    print 0, s
    exit()
else:
    find_recur(s)
    print(len(C))
    print " ".join(map(str, C))