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.
class solution:
    def __init__(self,n1,n2,n3):
        self.n1 = n1
        self.n2 = n2
        self.n3 = n3
        self.answer = 0
    def caluclation(self):
        for i in range(n1):
            if int(n2[i]) >= int(n3[i]):
                tar = int(n3[i])
                sm = int(n2[i])
            else:
                tar = int(n2[i])
                sm = int(n3[i])
            if abs(int(n2[i]) - int(n3[i])) > tar + 10 - sm:
                self.answer = self.answer + tar + 10 - sm
                
            else:
                self.answer = self.answer + abs(int(n2[i]) - int(n3[i]))
                



n1 = int(input())
n2 = input()
n3 = input()
root = solution(n1,n2,n3)
root.caluclation()
print(root.answer)
