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:
def power(a, b):
    if b == 0:
        return 1
    elif b % 2 == 0:
        a_1 = a * a
        return pow(a_1, b // 2)
    return a * pow(a, b - 1)


def k_fib(s, k):
    result = [0]
    if k >= 40:
        fibs = [power(2, i) for i in range(40)]
    else:
        fibs = [0] * (k - 1) + [1]
        while fibs[len(fibs) - 1] < s:
            fibs.append(sum(fibs[-k:]))
    while s > 0:
        if fibs[len(fibs) - 1] <= s:
            result.append(fibs[len(fibs) - 1])
            s -= fibs[len(fibs) - 1]
        fibs.pop()
    return result


S, K = [int(j) for j in input().split()]
print(len(k_fib(S, K)))
print(*k_fib(S, K))