/**
 * This solution implicitly builds a binary tree by
 * removing the nodes from lowest to highest level and
 * adding their worth and costs to their neighbours
 */

#include <algorithm>
#include <iostream>
#include <set>
#include <vector>

using namespace std;

typedef struct Node {
	Node() : left(nullptr),l_worth(0),
		right(nullptr), r_worth(0) {}

	Node *left;
	long long l_worth;
	Node *right;
	long long r_worth;
	long long level;
	long long req_lvl;
} Node;

int main() {
	size_t n;
	cin >> n;
	vector<Node> nodes(n);
	vector<long long> levels(n);
	long long highest_level = 0;
	size_t highest_level_index = 0;
	for(size_t i = 0; i < n; ++i) {
		long long level;
		cin >> level;
		levels[i] = level;
		if(level > highest_level) {
			highest_level_index = i;
			highest_level = level;
		}
	}

	size_t j = highest_level_index;
	vector<pair<long long, Node *>> q(n);
	for(size_t i = 0; i < n; ++i) {
		nodes[i].level = levels[j];
		nodes[i].req_lvl = nodes[i].level;
		q[i] = make_pair( nodes[i].level, &nodes[i] );
		j = (j + 1) % n;
	}

	Node empty;
	empty.right = &nodes[0]; empty.left = &nodes[n - 1];
	nodes[0].left = &empty; nodes[n - 1].right = &empty;
	for(size_t i = 0; i < n - 1; ++i) {
		nodes[i].right = &nodes[i + 1];
		nodes[i + 1].left = &nodes[i];
	}

	sort(q.begin(), q.end());
	for( const auto &p : q ) {
		const Node *me = p.second;
		Node *left = p.second->left;
		Node *right = p.second->right;
		long long worth = me->l_worth + me->level + me->r_worth;
		// Blindly overwriting works since we will only visit
		// the neighbours when it is cheapest to do so =>
		// worth can only increase, req_level can only decrease
		// (we still need std::min, since req_level comes from 2 sides)
		left->r_worth = worth;
		left->req_lvl = min( left->req_lvl,
				max( left->level - worth, me->req_lvl ) );
		right->l_worth = worth;
		right->req_lvl = min( right->req_lvl,
				max( right->level - worth, me->req_lvl ) );
		// Remove self
		left->right = right;
		right->left = left;
	}

	cout << q.back().second->req_lvl << endl;

	return 0;
}