//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using pli = pair<ll,int>;
#define AMARI 998244353
//#define AMARI 1000000007
#define el '\n'
#define El '\n'
#define YESNO(x) ((x) ? "Yes" : "No")
#define YES YESNO(true)
#define NO YESNO(false)
#define REV_PRIORITY_QUEUE(tp) priority_queue<tp,vector<tp>,greater<tp>>
#define EXIT_ANS(x) {cout << (x) << '\n'; return;}
template <typename T> void inline SORT(vector<T> &v){sort(v.begin(),v.end()); return;}
template <typename T> void inline VEC_UNIQ(vector<T> &v){sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); return;}
template <typename T> T inline MAX(vector<T> &v){return *max_element(v.begin(),v.end());}
template <typename T> T inline MIN(vector<T> &v){return *min_element(v.begin(),v.end());}
template <typename T> T inline SUM(vector<T> &v){T ans = 0; for(int i = 0; i < (int)v.size(); i++)ans += v[i]; return ans;}
void inline TEST(void){cerr << "TEST" << endl; return;}
vector<vector<int>> inline get_graph(int n,int m = -1,bool direct = false){
    if(m == -1)m = n - 1;
    vector<vector<int>> g(n);
    while(m--){
        int u,v;
        cin >> u >> v;
        u--; v--;
        g[u].push_back(v);
        if(!direct)g[v].push_back(u);
    }
    return g;
}


#define MULTI_TEST_CASE false
void solve(void){
    //問題を見たらまず「この問題設定から言えること」をいっぱい言う
    //よりシンプルな問題に言い換えられたら、言い換えた先の問題を自然言語ではっきりと書く
    //複数の解法のアイデアを思いついた時は全部メモしておく
    //g++ -D_GLIBCXX_DEBUG -Wall -O2 XXX.cpp -o o
    
    int n;
    cin >> n;
    vector<ll> a(n);
    for(int i = 0; i < n; i++)cin >> a[i];
    SORT(a);

    int ad = 0,mn = 0,mx = 0;
    for(int i = 0; i < n; i++){
        string s;
        cin >> s;
        if(s == "add")ad++;
        if(s == "min")mn++;
        if(s == "max")mx++;
    }
    if(mn == 0 && mx == 0){
        EXIT_ANS(SUM(a));
        return;
    }
    if(mx == 0 && ad == 0){
        EXIT_ANS(0);
        return;
    }
    if(ad == 0 && mn == 0){
        EXIT_ANS(MAX(a));
        return;
    }

    if(ad == 0){
        EXIT_ANS(a[mx - 1]);
        return;
    }

    if(mx == 0){
        ll ans1 = a[ad];
        ll ans2 = 0;
        for(int i = 0; i < ad; i++)ans2 += a[i];
        EXIT_ANS(min(ans1,ans2));
        return;
    }

    if(mn == 0){
        ll ans = a[ad + mx - 1];;
        for(int i = 0; i < ad - 1; i++){
            ans += a[i];
        }
        ans = max(ans,a[ad + mx - 1]);
        EXIT_ANS(ans);
        return;
    }
    assert(0);
    //cerr << ad << ' ' << mn << ' ' << mx << el;

    //後ろ mn 項を削って、mn == 0 の場合と同じような挙動になる？
    ll ans = a[ad + mx - 1];
    for(int i = 0; i < ad - 1; i++){
        ans += a[i];
    }
    ans = min(ans,a.back());
    cout << ans << El;
    return;
}

void calc(void){
    return;
}

signed main(void){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    calc();
    int t = 1;
    if(MULTI_TEST_CASE)cin >> t;
    while(t--){
        solve();
    }
    return 0;
}