Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.

The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
The first line contains a single integer *n* (1 ≤ *n* ≤ 1000) — the number of disks on the combination lock.

The second line contains a string of *n* digits — the original state of the disks.

The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
# def pick_peaks(arr):
# 	pos = []
# 	peaks = []
# 	i = 1
# 	while(i < (len(arr) - 1)):
# 		if (i == 1 and arr[i] == arr[0]):
# 			i = i + 1
# 			continue
# 		if (arr[i] > arr[i - 1] and arr[i] > arr[i + 1]):
# 			peaks.append(arr[i])
# 			pos.append(i)
# 		elif (arr[i] == arr[i + 1] and arr[i] > arr[i - 1]):
# 			j = i + 1
# 			while(j < len(arr)):
# 				if(arr[j] != arr[i]):
# 					break
# 				j = j + 1
# 			if (j != len(arr)):
# 				if (arr[j] < arr[i]):
# 					peaks.append(arr[i])
# 					pos.append(i)
# 					i = j
# 					continue
# 		elif(arr[i] == arr[i + 1] and arr[i] == arr[i - 1]):
# 			flag1 = 0
# 			flag2 = 0
# 			u = 0
# 			j = i + 1
# 			while(j < len(arr)):
# 				if(arr[j] != arr[i]):
# 					break
# 				j = j + 1
# 			if (j != len(arr)):
# 				if (arr[j] < arr[i]):
# 					flag1 = 1
# 			u = j
# 			j = 0
# 			while(j < i):
# 				if(arr[j] != arr[i]):
# 					break
# 				j = j + 1
# 			if (arr[j] < arr[i]):
# 				flag2 = 1
# 			if (flag1 and flag2):
# 				peaks.append(arr[i])
# 				pos.append(i)
# 				i = u
# 				continue
# 		elif(arr[i] == arr[i - 1] and arr[i] > arr[i + 1]):
# 			j = 0
# 			while(j < i):
# 				if(arr[j] != arr[i]):
# 					break
# 				j = j + 1
# 			if (arr[j] < arr[i]):
# 				peaks.append(arr[i])
# 				pos.append(i)
# 		i = i + 1
# 	return {'pos' : pos, 'peaks' : peaks}
def solve(n, orginal, secret):
	ans = 0
	for i in range(n):
		digit1 = (int)(orginal[i])
		digit2 = (int)(secret[i])
		if (abs(digit2 - digit1) > 5):
			ans += abs(9 - max(digit1, digit2)) + min(digit2, digit1) + 1
		else:
			ans += abs(digit2 - digit1)
	print(ans)

n = (int)(input())
orginal = input()
secret = input()
solve(n, orginal, secret)
