#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, l, s, m, a, b, c;
  cin >> n;
  string s1, s2;
  cin >> s1;
  cin >> s2;
  int p, q;
  n = s1.size();
  n = s2.size();
  int cot = 0;
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] > s2[i]) {
      a = abs(s2[i] - s1[i]);
      b = s2[i] + 10;
      c = b - s1[i];
      if (c < a) {
        cot = cot + c;
      } else {
        cot = cot + a;
      }
    } else {
      k = abs(s1[i] - s2[i]);
      l = s1[i] + 10;
      m = l - s2[i];
      if (m < k) {
        cot = cot + m;
      } else {
        cot = cot + k;
      }
    }
  }
  cout << cot << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

<image>

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?

Input

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.

Output

Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.

Examples

Input

5
82195
64723


Output

13

Note

In the sample he needs 13 moves:

  * 1 disk: <image>
  * 2 disk: <image>
  * 3 disk: <image>
  * 4 disk: <image>
  * 5 disk: <image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, l, s, m, a, b, c;
  cin >> n;
  string s1, s2;
  cin >> s1;
  cin >> s2;
  int p, q;
  n = s1.size();
  n = s2.size();
  int cot = 0;
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] > s2[i]) {
      a = abs(s2[i] - s1[i]);
      b = s2[i] + 10;
      c = b - s1[i];
      if (c < a) {
        cot = cot + c;
      } else {
        cot = cot + a;
      }
    } else {
      k = abs(s1[i] - s2[i]);
      l = s1[i] + 10;
      m = l - s2[i];
      if (m < k) {
        cot = cot + m;
      } else {
        cot = cot + k;
      }
    }
  }
  cout << cot << endl;
}

```