#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const ll INF = 1e18;
const int MX = 100000;
const int MOD = (119 << 23) + 1, root = 3; // = 998244353
// For p < 2^30 there is also e.g. (5 << 25, 3), (7 << 26, 3),
// (479 << 21, 3) and (483 << 21, 5). The last two are > 10^9.
ll po (ll b, ll p) { return !p?1:po(b*b%MOD,p/2)*(p&1?b:1)%MOD; }
ll inv (ll b) { return po(b,MOD-2); }
int ad(int a, int b) { a += b; if (a >= MOD) a -= MOD; return a; }
int sub(int a, int b) { return (a-b+MOD)%MOD; }
int mul(int a, int b) { return (ll)a*b%MOD; }
int AD(int& a, int b) { return a = ad(a,b); }
int SUB(int& a, int b) { return a = sub(a,b); }
int MUL(int& a, int b) { return a = mul(a,b); }
int L[1<<18], R[1<<18], tmp[1<<18];
namespace NTT {
int get(int s) {
return s > 1 ? 32 - __builtin_clz(s - 1) : 0;
}
int roots[1<<18];
void ntt(int *a, bool f = 0) { // clearly not own
int N = 1<<18, P = MOD;
int i, j, k, x, y, z;
j = 0;
for (i = 1; i < N; i++) {
for (k = N >> 1; j >= k; k >>= 1) j -= k;
j += k;
if (i < j) {
k = a[i];
a[i] = a[j];
a[j] = k;
}
}
for (i = 1; i < N; i <<= 1) {
x = po(3, P / i >> 1);
for (j = 0; j < N; j += i << 1) {
y = 1;
for (k = 0; k < i; k++) {
z = (long long)a[i | j | k] * y % P;
a[i | j | k] = a[j | k] - z;
if (a[i | j | k] < 0) a[i | j | k] += P;
a[j | k] += z;
if (a[j | k] >= P) a[j | k] -= P;
y = (long long)y * x % P;
}
}
}
}
/*
void ntt(int* a) {
int n = 1<<18, x = 18;
roots[0] = 1, roots[1] = po(root,(MOD-1)/n);
FOR(i,2,n) roots[i] = mul(roots[i-1],roots[1]);
array<int,1<<18> res, RES;
F0R(i,1<<18) res[i] = a[i];
FOR(i,1,x+1) {
int inc = n>>i;
F0R(j,inc) for (int k = 0; k < n; k += inc) {
int t = 2*k+j; if (t >= n) t -= n;
RES[k+j] = ad(res[t],mul(roots[k],res[t+inc]));
}
swap(res,RES);
}
F0R(i,1<<18) a[i] = res[i];
}*/
void ntt_rev(int* a) {
ntt(a);
int in = inv(1<<18);
F0R(i,1<<18) MUL(a[i],in);
reverse(a+1,a+(1<<18));
}
void conv() {
ntt(L), ntt(R);
F0R(i,1<<18) tmp[i] = mul(L[i],R[i]);
ntt_rev(tmp);
//F0R(i,1<<18) cout << tmp[i] << " ";
//cout << "\n";
}
}
ll ans = INF;
int res[10][MX], res2[10][MX], RES[MX], RES2[MX];
int K, B[MX], digYes[MX], digNo[MX];
int match(int x, int y) {
while (x) {
if (x % 10 == y) return 1;
x /= 10;
}
return 0;
}
ll tri(int x, int y) {
if (x == 0) return y;
if (x == 1) x ^= 2;
ll t = 0;
FOR(i,1,10) if (x&(1<<i)) {
t = i;
x ^= 1<<i;
break;
}
F0R(i,10) if (x&(1<<i)) {
t = 10*t+i;
x ^= 1<<i;
}
return t;
}
ll solve(pi a, pi b) { // what about leading zeroes??
ll t = INF;
F0R(i,9) {
int A = min(a.f,a.f^(1<<i));
int B = min(b.f,b.f^(1<<(i+1)));
t = min(t,10*tri(A|B,(a.f&1) || (i == 0 && a.s))+i);
A = min(A,A^(1<<9));
B = min(B,B^(1<<0));
t = min(t,100*tri(A|B,a.s)+10*i+9);
}
return MX*t;
}
pi get0(int st) {
pi ret = {0,0};
F0R(i,10) if (res[i][st] == 1) ret.f ^= 1<<i;
if (RES[st] == 1) ret.s = 1;
return ret;
}
pi get1(int st) {
pi ret = {0,0};
F0R(i,10) if (res2[i][st] == 1) ret.f ^= 1<<i;
if (RES2[st] == 1) ret.s = 1;
return ret;
}
ll test(int st) {
pi a = get0(st), b = get1(st);
// if (st == 0) cout << a.f << " " << a.s << "\n";
ll ret = st+solve(a,b);
return ret;
}
void gen() {
ios_base::sync_with_stdio(0); cin.tie(0);
FOR(i,1,MX) digNo[i] = digNo[i/10]|(1<<(i%10));
F0R(i,MX) {
digYes[i] = digNo[i];
if (i < MX/10) digYes[i] |= 1;
}
cin >> K;
F0R(i,K) cin >> B[i];
}
void oops(int x) {
memset(L,0,sizeof L);
memset(R,0,sizeof R);
F0R(i,MX) L[i] = !(digYes[i]&(1<<x));
F0R(i,K) R[MX-i] = (B[i] == x);
NTT::conv();
F0R(i,MX) if (tmp[i]) res2[x][i] = 1;
F0R(i,MX) if (tmp[i+MX]) res[x][i] = 1;
}
void OOPS(int x = 0) {
memset(L,0,sizeof L);
memset(R,0,sizeof R);
F0R(i,MX) L[i] = !(digNo[i]&(1<<x));
F0R(i,K) R[MX-i] = (B[i] == x);
NTT::conv();
F0R(i,MX) if (tmp[i]) RES2[i] = 1;
F0R(i,MX) if (tmp[i+MX]) RES[i] = 1;
}
int main() {
gen();
F0R(i,10) oops(i);
OOPS();
F0R(i,MX) ans = min(ans,test(i));
cout << ans;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
