#include <iostream>
#include <fstream>
#include <algorithm>

using namespace std;

int v[100005];
long long dp[100005];

// Functie DP care calculeaza maximul pe intervalul [st, dr]
long long solve(int st, int dr) {
    if (st > dr) return 0;
    int len = dr - st + 1;
    if (len < 2) return 0;

    // Resetam dp pentru portiunea folosita
    // dp[i] va reprezenta maximul pana la elementul st + i
    for (int i = 0; i < len; i++) dp[i] = 0;

    for (int i = 1; i < len; i++) {
        // Nu luam o pereche care se termina la elementul i
        dp[i] = dp[i - 1];

        // Luam perechea (i-1, i)
        long long current_pair = (long long)v[st + i - 1] + v[st + i];

        if (i >= 3) {
            dp[i] = max(dp[i], current_pair + dp[i - 3]);
        } else {
            dp[i] = max(dp[i], current_pair);
        }
    }
    return dp[len - 1];
}

int main() {
    ifstream fin("oo.in");
    ofstream fout("oo.out");

    int n;
    if (!(fin >> n)) return 0;

    for (int i = 0; i < n; i++) {
        fin >> v[i];
    }

    if (n == 2) {
        fout << (long long)v[0] + v[1] << endl;
        return 0;
    }

    long long max_total = 0;

    // Cazul 1: Ignoram conexiunea N-1 cu 0 (Liniar standard)
    max_total = max(max_total, solve(0, n - 1));

    // Cazul 2: Fortam perechea (v[n-1], v[0])
    // Blocheaza vecinii: v[n-2] si v[1]
    // Ramane liber: [2, n-3]
    max_total = max(max_total, (long long)v[n - 1] + v[0] + solve(2, n - 3));

    // Cazul 3: Fortam perechea (v[0], v[1])
    // Blocheaza vecinii: v[n-1] si v[2]
    // Ramane liber: [3, n-2]
    max_total = max(max_total, (long long)v[0] + v[1] + solve(3, n - 2));

    // Cazul 4: Fortam perechea (v[1], v[2])
    // Blocheaza vecinii: v[0] si v[3]
    // Ramane liber: [4, n-1]
    max_total = max(max_total, (long long)v[1] + v[2] + solve(4, n - 1));

    fout << max_total << endl;

    return 0;
}