/**
Avem maximul la un pas. Pt a alege un nou element, Trebuie sa extindem o secventa. Tinem un set cu extinderile.
**/
#include <iostream>
#include <set>
#define dim 200000
#define ppp pair <long long,pair <int,int>>
#define cost first
#define l second.first
#define r second.second
using namespace std;
set <ppp> s;
set <ppp>::iterator it;
ppp pos[dim+1];
int n,v[dim+1];
long long ss[dim+1];
void init ()
{
cin>>n;
for (int i=1; i<=n; i++)
{
cin>>v[i];
if (i==1)
ss[1]=v[1];
else
ss[i]=ss[i-2]+v[i];
pos[i]=make_pair (v[i],make_pair (i,i));
s.insert (make_pair (v[i],make_pair (i,i)));
}
}
long long get_cost (int st,int dr)
{
return (ss[dr]-ss[st]-v[dr])-(ss[dr-1]-ss[st+1]+v[st+1]);
}
void solve ()
{
long long sol=0;
for (int i=1; i<=(n+1)/2; i++)
{
it=s.end (),it--;
ppp elem=*it;
sol+=elem.first;
elem.l--;
elem.r++;
s.erase (it);
if (s.find (pos[elem.l])!=s.end ())
s.erase (s.find (pos[elem.l]));
if (s.find (pos[elem.r])!=s.end ())
s.erase (s.find (pos[elem.r]));
ppp nelem=make_pair (pos[elem.l].cost+pos[elem.r].cost+get_cost (elem.l,elem.r),make_pair (pos[elem.l].l,pos[elem.r].r));
if (nelem.l>0&&nelem.r<=n)
s.insert (nelem);
pos[nelem.l]=nelem,pos[nelem.r]=nelem;
cout<<sol<<"\n";
}
}
int main ()
{
init ();
solve ();
return 0;
}
