#include <bits/stdc++.h>
#pragma GCC optimize ("Ofast,unroll-loops")
using namespace std;
typedef long long ll;
typedef pair<int, int> pp;
#define rep(i,l,r) for(int i = (l); i < (r); i++)
#define per(i,r,l) for(int i = (r); i >= (l); i--)
#define sz(x) (int)x.size()
#define ff first
#define ss secounf
#define all(x) begin(x), end(x)
#define pb push_back
const ll mod = 1e9+7, maxn = 5e2+1, inf = ll(1e9)+5;
ll slv(vector<int> a, bool can = true){
bool ok = false;
for(int c: a) ok |= c > 0;
if(!ok) return 0;
if(sz(a) == 1){
int x = a[0], res = 0, mn = 0;
if(x&1){
mn = 1;
per(i,10,1) if(x>>i&1) mn = i;
res = mn*10;
}
rep(i,mn+1,10) if(x>>i&1) res = res*10 + i;
return res;
}
// cout << sz(a) << endl;
ll ans = 10987654321000000ll;
rep(d,0,10){
if(d == 9 && !can) break;
vector<int> b{0};
bool has_zero = false;
int cr = d;
rep(i,0,sz(a)){
if(a[i]>>cr&1) b.back() |= a[i]^(1<<cr);
else b.back() |= a[i];
if(!cr && (a[i]&1)) has_zero = true;
cr++;
if(cr == 10) {
cr = 0;
if(i - sz(a) + 1) b.pb(0);
}
}
// cout << sz(b) << '\n';
ll res = slv(b, d != 9 || sz(a) > 2)*10 + d;
if(!res && has_zero) res = 10;
ans = min(ans, res);
}
return ans;
}
int main(){
cin.tie(0) -> sync_with_stdio(0);
int n; cin >> n;
vector<int> a(n);
rep(i,0,n){
cin >> a[i], a[i] = 1<<a[i];
}
cout << slv(a) << '\n';
return 0;
}
