/*
Baltic 2014 Sequence
- Let each number K be represented as (X)Y, where X = K / 10 and Y = K % 10
- Notice that now if we remove the last digit from each number, we have
(X), (X), ..., (X+1), (X+1), ...
- i.e. we have contiguous segments with the same prefix
- We can thus merge them together and solve recursively
- For each number, consider the set of digits that it must contain
- Initially, this set is simply the given digit for each number
- If we know the last digit of some number, we can simply remove that digit from the set
- When we only have 1 number, the answer is simply the smallest non-zero number
with no leading zeroes we can make from the set
- Our solution is thus simply to brute force over each last digit and solve
the subproblems recursively
- There are N/10 subproblems because we group numbers in groups of 10
- Also, to efficiently represent the sets of digits, we use bitmasking since then
we can just add masks for set unions
- Complexity: O(N log N)
*/
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC target("sse4,avx2,fma,avx")
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
ll ans = 1e16, pw[18];
void solve(vector<short> sections, ll curr = 0, int level = 0, bool had_zero = false) {
if (curr > ans) return;
int n = sections.size();
if (n == 1) {
int mask = sections[0];
if (!mask) {
if (!curr || had_zero) curr += pw[level];
} else if (mask == 1) curr += pw[level + 1];
else {
if (mask & 1) {
int mn = 10;
for (int i = 9; i; i--) {
if (mask & (1 << i)) mn = i;
}
for (int i = 9; i; i--) {
if ((mask & (1 << i)) && i != mn) curr += i * pw[level++];
}
curr += mn * pw[++level];
} else {
for (int i = 9; i; i--) {
if (mask & (1 << i)) curr += i * pw[level++];
}
}
}
ans = min(ans, curr);
return;
}
FOR(i, 0, 10) {
vector<short> merged;
int ins = 0;
bool zero = false;
int piv = i;
FOR(j, 0, n) {
zero |= (piv == 0 && (sections[j] & 1));
ins |= (sections[j] & (~(1 << piv)));
piv++;
if (piv == 10) {
merged.push_back(ins);
ins = piv = 0;
}
}
if (piv) merged.push_back(ins);
solve(merged, curr + pw[level] * i, level + 1, zero);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector<short> sections(n);
FOR(i, 0, n) {
int x;
cin >> x;
sections[i] = 1 << x;
}
pw[0] = 1;
FOR(i, 1, 18) pw[i] = 10 * pw[i - 1];
solve(sections);
cout << ans;
return 0;
}
