Problem:
Problem C: Dial Lock

A dial lock is a kind of lock which has some dials with printed numbers. It has a special sequence of
numbers, namely an unlocking sequence, to be opened.

You are working at a manufacturer of dial locks. Your job is to verify that every manufactured lock is
unlocked by its unlocking sequence. In other words, you have to rotate many dials of many many locks.

It’s a very hard and boring task. You want to reduce time to open the locks.
It’s a good idea to rotate multiple dials at one time. It is, however, a difficult problem to find steps to open
a given lock with the fewest rotations. So you decided to write a program to find such steps for given
initial and unlocking sequences.

Your company’s dial locks are composed of vertically stacked k (1 ≤ k ≤ 10) cylindrical dials. Every dial
has 10 numbers, from 0 to 9, along the side of the cylindrical shape from the left to the right in sequence.
The neighbor on the right of 9 is 0.

A dial points one number at a certain position. If you rotate a dial to the left by i digits, the dial newly
points the i-th right number. In contrast, if you rotate a dial to the right by i digits, it points the i-th left
number. For example, if you rotate a dial pointing 8 to the left by 3 digits, the dial newly points 1.

You can rotate more than one adjacent dial at one time. For example, consider a lock with 5 dials. You
can rotate just the 2nd dial. You can rotate the 3rd, 4th and 5th dials at the same time. But you cannot
rotate the 1st and 3rd dials at one time without rotating the 2nd dial. When you rotate multiple dials, you
have to rotate them to the same direction by the same digits.

Your program is to calculate the fewest number of rotations to unlock, for given initial and unlocking
sequences. Rotating one or more adjacent dials to the same direction by the same digits is counted as one
rotation.

Input

The input consists of multiple datasets.

Each datset consists of two lines. The first line contains an integer k. The second lines contain two strings,
separated by a space, which indicate the initial and unlocking sequences.

The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed.

Output

For each dataset, print the minimum number of rotations in one line.

Sample Input

4
1357 4680
6
777777 003330
0

Output for the Sample Input

1
2

Solution in C++14 (judge verdict: Accepted)

#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  while(true) {
    int n;
    cin >> n;
    if(!n) break;
    string s, t;
    cin >> s >> t;

    int ans = LLINF;
    function<void(int,int,string)> dfs = [&](int idx, int cost, string str) {
      if(idx == n || str == t) { chmin(ans, cost); return; }
      // idx番目をstr[idx]->t[idx]にもっていくのにかかるコスト
      int diff = (t[idx]-str[idx]+10)%10;
      // [idx, i] を回転させる
      FOR(i, idx, str.size()) {
        str[i] = ((str[i]-'0') + diff)%10 + '0';
        dfs(idx+1, cost+(diff==0?0:1), str);
      }
    };

    dfs(0, 0, s);
    cout << ans << endl;
  }

  return 0;
}