Array-Inversion-Count

from bisect import bisect_left
def merge(l):
    if len(l) < 2:
        return [0,l]
    else:
        a, left = merge(l[:len(l)//2])
        b, right = merge(l[len(l)//2:])

        ret = a+b
        for i in range(len(left)):
            ret += bisect_left(right, left[i])
        return [ret, sorted(left+right)]
def solution(A):
    ret = merge(A)[0]
    return ret if ret <= 1000000000 else -1