human
Find the value of N XOR 'ed to itself K times | Function to return n ^ n ^ ... k times.
If k is odd the answer is the number itself.
Else the answer is 0.
Driver code"
How can the above be solved in Python?

Can you provide a code to find the value of N XOR 'ed to itself K times and the other steps required?
assistant
#Here is code to find the value of N XOR 'ed to itself K times and the other steps required.
 def xorK ( n , k ) :
	if ( k % 2 == 1 ) :
		return n
	return 0
n = 123
k = 3
print ( xorK ( n , k ) )