// NOTE: this implementation may contain errors ^^'
//
// solution in n*log(n)^2
// idee:
// we determine the minimum cost for each floor (n times)
//   for the ith floor we start with level in[i]+1.
//   Therefore we can kill all enemies in the range [i, i+1)
//   now we determine two person outside of the current range which are: (takes log(n))
//     - as close as possible
//     - have a level whose MSB is >= the MSB of our level
//   if we can't kill any of those two we increase our starting level to be able to do so
//   We can kill the one with lower level
//   => our MSB increases (this can only happen log(n) times)
#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 bits = 46;
constexpr ll limN = 500007;
constexpr ll inf = 0x3FFFFFFFFFFFFFF;

vector<ll> arr[bits];
ll in[limN], pref[limN];

void add(ll x, ll i) {
	in[i] = x;
	pref[i + 1] = pref[i] + in[i];
	for (ll j = 0; j < bits; j++) {
		if ((1ll << j) <= in[i]) arr[j].push_back(i);
	}
}

// is __lg(0) defined?
ll mylog(ll x) {
	if (x == 0) return 0;
	else return __lg(x);
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	ll n;
	cin >> n;
	add(inf, 0);
	{
		deque<ll> tmp(n);
		for (ll i = 0; i < n; i++) cin >> tmp[i];
		ll ma = *max_element(all(tmp));
		while (tmp[0] != ma) {
			tmp.push_back(tmp[0]);
			tmp.pop_front();
		}
		for (ll i = 0; i < n; i++) add(tmp[i], i+1);
	}
	add(inf, n+1);
	ll res = inf;
	for (ll i = 1; i <= n; i++) {
		ll cur = in[i];
		ll l = i;
		ll r = i + 1;
		while (l > 1 || r < n + 1) {
			ll tmp = cur + pref[r] - pref[l];
			ll tmpBits = min(bits-1, mylog(tmp));
			r = *lower_bound(all(arr[tmpBits]), r);
			l = *prev(upper_bound(all(arr[tmpBits]), l)) + 1;
			tmp = cur + pref[r] - pref[l];
			if (l > 1 || r < n + 1) {
				cur += max(0ll, min(in[l - 1], in[r]) - tmp);
				in[l-1] < in[r] ? l--: r++;
			}
		}
		res = min(res, cur);
	}
	cout << res << endl;
}