# Python3 implementation of the approach 

# Function to return the 
# reverse of a number 
def reverse(n): 
	rev = 0
	while (n != 0): 
		rev = (rev * 10) + (n % 10) 
		n //= 10
	return rev 

# Function to find the sum of the odd 
# and even positioned digits in a number 
def getSum(n): 

	n = reverse(n) 
	sumOdd = 0
	sumEven = 0
	c = 1

	while (n != 0): 

		# If c is even number then it means 
		# digit extracted is at even place 
		if (c % 2 == 0): 
			sumEven += n % 10
		else: 
			sumOdd += n % 10
		n //= 10
		c += 1

	print("Sum odd =", sumOdd) 
	print("Sum even =", sumEven) 

# Driver code 
n = 457892
getSum(n) 

