#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #ifdef LOCAL # include "debug_print.hpp" # define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__) #else # define debug(...) (static_cast(0)) #endif using namespace std; #define rep(i,n) for(int i=0; i<(n); i++) #define rrep(i,n) for(int i=(n)-1; i>=0; i--) #define FOR(i,a,b) for(int i=(a); i<(b); i++) #define RFOR(i,a,b) for(int i=(b-1); i>=(a); i--) #define ALL(v) v.begin(), v.end() #define RALL(v) v.rbegin(), v.rend() #define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() ); #define pb push_back using ll = long long; using D = double; using LD = long double; using P = pair; template using PQ = priority_queue>; template using minPQ = priority_queue, greater>; templatebool chmax(T &a, const T &b) { if (abool chmin(T &a, const T &b) { if (b ostream &operator<<(ostream &os, const pair &p) { os << p.first << " " << p.second; return os; } template istream &operator>>(istream &is, pair &p) { is >> p.first >> p.second; return is; } template ostream &operator<<(ostream &os, const vector &v) { int s = (int)v.size(); for (int i = 0; i < s; i++) os << (i ? " " : "") << v[i]; return os; } template istream &operator>>(istream &is, vector &v) { for (auto &x : v) is >> x; return is; } void in() {} template void in(T &t, U &...u) { cin >> t; in(u...); } void out() { cout << "\n"; } template void out(const T &t, const U &...u) { cout << t; if (sizeof...(u)) cout << sep; out(u...); } void outr() {} template void outr(const T &t, const U &...u) { cout << t; outr(u...); } int main(){ ios_base::sync_with_stdio(false); cin.tie(nullptr); string s; in(s); int n = s.size(); vector dp((1 << n), -1); auto rec = [&] (auto rec, int t) -> ll{ if(dp[t] != -1) return dp[t]; if(__builtin_popcount(t) < 3) return dp[t] = 0; ll res = 0; rep(i,n)FOR(j,i+1,n)FOR(k,j+1,n){ if((t >> i)&1 && (t >> j)&1 && (t >> k)&1 && s[i] != '0' && s[j] == s[k] && s[i] != s[j]){ ll nxt = t & ~(1 << i) & ~(1 << j) & ~(1 << k); chmax(res, rec(rec, nxt) + (s[i]-'0')*100 + (s[j]-'0')*11); } } return dp[t] = res; }; out(rec(rec,(1 << n)-1)); }