【CodeForces 1260E --- Tournament】

You are organizing a boxing tournament, where n boxers will participate (n is a power of 2), and your friend is one of them. All boxers have different strength from 1 to n, and boxer i wins in the match against boxer j if and only if i is stronger than j.

The tournament will be organized as follows: n boxers will be divided into pairs; the loser in each pair leaves the tournament, and n2 winners advance to the next stage, where they are divided into pairs again, and the winners in all pairs advance to the next stage, and so on, until only one boxer remains (who is declared the winner).

Your friend really wants to win the tournament, but he may be not the strongest boxer. To help your friend win the tournament, you may bribe his opponents: if your friend is fighting with a boxer you have bribed, your friend wins even if his strength is lower.

Furthermore, during each stage you distribute the boxers into pairs as you wish.

The boxer with strength i can be bribed if you pay him ai dollars. What is the minimum number of dollars you have to spend to make your friend win the tournament, provided that you arrange the boxers into pairs during each stage as you wish?

Input

The first line contains one integer n (2≤n≤218) — the number of boxers. n is a power of 2.

The second line contains n integers a1, a2, …, an, where ai is the number of dollars you have to pay if you want to bribe the boxer with strength i. Exactly one of ai is equal to −1 — it means that the boxer with strength i is your friend. All other values are in the range [1,109].

Output

Print one integer — the minimum number of dollars you have to pay so your friend wins.

Sample Input

4
3 9 1 -1

Sample Output

0

AC代码：

#include 
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
using ll = long long;
const int MAXN = (1<<18)+5;
ll arr[MAXN];
bool b[MAXN];

int main()
{
    SIS;
    int n,x=1;
    cin >> n;
    for(int i=1;i<=n;i++) cin >> arr[i];
    while(x<=n) b[x]=true,x<<=1;
    multiset<ll> s;
    ll ans=0;
    for(int i=n;i>0;i--)
    {
        if(arr[i]==-1) break;
        s.insert(arr[i]);
        if(b[i]) 
        {
            ans+=*s.begin();
            s.erase(s.begin());
        }
    }
    cout << ans << endl;
    return 0;
}