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: introductory

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: # Check input whether it is an integer
     return pow(int(input)+1,2) # Return the next square if sq is a square
    return -1 # Return -1 otherwise


import math


def find_next_square(sq):
    """Return the next square if sq is a square, -1 otherwise"""
    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

from math import sqrt
def find_next_square(sq):

    if int(sqrt(sq)) == sqrt(sq):
        return (sqrt(sq) + 1) ** 2
    else:
        return -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