Write a function name `nextPerfectSquare` that returns the first perfect square that is greater than its integer argument. A `perfect square` is a integer that is equal to some integer squared. For example 16 is a perfect square because `16=4*4`. 
```
example

n   next perfect sqare

6    9
36   49 
0    1
-5   0 
```
```if-not:csharp
caution! the largest number tested is closer to `Number.MAX_SAFE_INTEGER`
```
```if:csharp
Caution! The largest number test is close to `Int64.MaxValue`
```

Difficulty: EASY
Starter code: def next_perfect_square(n):
	

Solutions:
def next_perfect_square(n):
	return n >= 0 and (int(n ** 0.5) + 1) ** 2


next_perfect_square = lambda n: (n >= 0 and int(n ** 0.5) + 1) ** 2


import math

def next_perfect_square(n):
	return math.ceil((n + 1) ** 0.5) ** 2 if n > -1 else 0


from bisect import bisect
squares = [i * i for i in range(32000)]

def next_perfect_square(n):
	return bisect(squares, n) ** 2


next_perfect_square = lambda n: (int(n ** 0.5) + 1) ** 2 if n >= 0 else 0


def next_perfect_square(n, i=0):
	while True:
		if i * i > n:
			return i * i
		i += 1


def next_perfect_square(n):
	if n < 0:
		return 0
	is_square = lambda s: int(s ** 0.5) ** 2 == s
	n += 1
	while not is_square(n):
		n += 1
	return n


from math import sqrt

def next_perfect_square(n):
	return 0 if n < 0 else (int(sqrt(n)) + 1) ** 2


from math import floor

def next_perfect_square(n):
	return (floor(n ** (1 / 2)) + 1) ** 2 if n > 0 else 0 if n != 0 else 1