You might know some pretty large perfect squares.  But what about the NEXT one?

Complete the `findNextSquare` method that finds the next integral perfect square after the one passed as a parameter.  Recall that an integral perfect square is an integer n such that sqrt(n) is also an integer.  

If the parameter is itself not a perfect square then `-1` should be returned.  You may assume the parameter is positive.


**Examples:**

```
findNextSquare(121) --> returns 144
findNextSquare(625) --> returns 676
findNextSquare(114) --> returns -1 since 114 is not a perfect
```

Difficulty: EASY
Starter code: def find_next_square(sq):
	

Solutions:
def find_next_square(sq):
	root = sq ** 0.5
	if root.is_integer():
		return (root + 1) ** 2
	return -1


def find_next_square(sq):
	x = sq ** 0.5
	return -1 if x % 1 else (x + 1) ** 2


import math

def find_next_square(sq):
	return (math.sqrt(sq) + 1) ** 2 if math.sqrt(sq).is_integer() else -1


from math import sqrt

def find_next_square(sq):
	return (sqrt(sq) + 1) ** 2 if sqrt(sq) % 1 == 0 else -1


import math

def find_next_square(sq):
	input = float(math.sqrt(sq))
	if input % 1 == 0:
		return pow(int(input) + 1, 2)
	return -1


import math

def find_next_square(sq):
	square = math.sqrt(float(sq))
	return -1 if not square.is_integer() else math.pow(square + 1, 2)


def find_next_square(sq):
	return (sq ** 0.5 + 1) ** 2 if int(sq ** 0.5) ** 2 == sq else -1


def find_next_square(sq):
	import math
	f = float(sq)
	root = math.sqrt(f)
	if root.is_integer():
		next_int = int(root) + 1
		next_square = next_int * next_int
		return next_square
	else:
		return -1


def find_next_square(sq):
	sqrt = sq ** 0.5
	if sqrt % 1 == 0:
		return (sqrt + 1) ** 2
	return -1