// binary search solution:
// for a fixed start level m, go through each level from left to right
//   - if we can currently beat that level gain the x levels.
//     after this try to beat as many levels as possible from the todo 
//     stack (beat the first value and gain the second as level)
//   - add a pair {x, x+current level} on a todo stack.
// if we can clear the todo stack we can beat all levels.
#include <bits/stdc++.h>
using namespace std;

#define all(x) begin(x), end(x)
#define sz(x) (ll)(x).size()

using ll = long long;
using ld = long double;

constexpr ll inf = 10000000000007;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	ll n;
	cin >> n;
	vector<ll> in(2*n);
	for (ll i = 0; i < n; i++) {
		cin >> in[i];
		in[i+n] = in[i];
	}
	
	auto ok = [&](ll m){
		vector<pair<ll, ll>> todo;
		ll cur = m;
		for (ll x : in) {
			if (x <= cur) {
				cur += x;
				while (!todo.empty() && todo.back().first <= cur) {
					cur += todo.back().second;
					todo.pop_back();
				}
			} else {
				todo.push_back({x, cur - m + x});
				cur = m;
			}
		}
		return todo.empty();
	};

	ll l = -1;
	ll r = inf;
	while (l + 1 < r) {
		ll m = (l+r)/2;
		if (ok(m)) {
			r = m;
		} else {
			l = m;
		}
	}
	cout << r << endl;
}