// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define pf push_front
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())
template<class T> using V = vector<T>;
using pi = pair<int, int>;
using vi = V<int>;
using vpi = V<pi>;
using ll = long long;
using pl = pair<ll, ll>;
using vpl = V<pl>;
using vl = V<ll>;
using db = long double;
template<class T> using pq = priority_queue<T, V<T>, greater<T>>;
const int MOD = 1e9 + 7;
const ll INFL = ll(1e17);
const int LG = 18;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vi A(N); for(auto& x : A) cin >> x;
for(int i = 0; i < N; i++) A.pb(A[i]);
vl P = {0}; for(int i = 0; i < 2*N; i++) P.pb(P.back() + A[i]);
auto qry = [&](int l, int r) {
return P[r+1] - P[l];
};
int M = (N + 1) / 2;
vi S;
for(int i = 0; (i+M-1) < 2*N; i++) S.pb(qry(i, i + M - 1));
multiset<int> MN;
int ans = 0;
for(int i = 0; i < sz(S); i++) {
MN.insert(S[i]);
if (i-M >= 0) MN.erase(MN.find(S[i-M]));
if (sz(MN) == M) ans = max(ans, *begin(MN));
}
cout << ans << nl;
exit(0-0);
}
