#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#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 MOD 1000000007
#define INF (1LL<<25)     //33554432
#define LLINF (1LL<<60)   //1152921504606846976
#define PI 3.14159265359
#define EPS 1e-12
//#define int ll

//二分累乗法
ll binpow(ll x, ll e)
{
  ll a = 1, p = x;

  while(e > 0) {
    if(e%2 == 0) {p = (p*p); e /= 2;}
    else {a = (a*p); e--;}
  }

  return a;
}

signed main(void)
{
  int n;
  string v[1010];
  cin >> n;
  REP(i, n) cin >> v[i];

  ll ans = -1;
  FOR(i, 2, 37) {
    REP(j, n) {
      ll ret = 0;
      bool flag = true;
      REP(k, v[j].size()) {
        if((i <= 9 && v[j][k]-'0' >= i)
          || (i <= 9 && v[j][k] >= 'A' && v[j][k] <= 'Z')
          || (i >= 10 && v[j][k] >= 'A' && v[j][k] <= 'Z' && v[j][k] >= 'A'+i-10)) {
          flag = false;
          break;
        }
        //cout << v[j].size()-k-1 << " " << binpow(i, v[j].size() - k - 1) << " " << (v[j][k] - '0') << endl;
        if(v[j][k] >= '0' && v[j][k] <= '9')
          ret += binpow(i, v[j].size() - k - 1) * (v[j][k] - '0');
        else
          ret += binpow(i, v[j].size() - k - 1) * (v[j][k] - 'A' + 10);
      }
      //cout << "i:" << i << " j:" << j << " ret:" << ret;
      if(flag) {
        if(ans == -1) ans = ret;
        else ans = min(ans, ret);
      }
      //cout << " ans:" << ans << endl;
    }
  }

  cout << ans << endl;

  return 0;
}