//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define int long long
typedef pair<int, int> T;
void solve(){
int n; cin >> n;
vector<int>a(n+2), L(n+2), R(n+2);
for (int i = 1; i<=n; i++) {
cin >> a[i];
}
iota(L.begin(), L.end(), 0);
iota(R.begin(), R.end(), 0);
auto cost = a;
priority_queue<T>pq;
vector<bool>beg(n+2, 1);
for (int i = 1; i<=n; i++) pq.push({a[i], i});
//niezmiennik --> konce przedzialow sa puste
//przedzial ma zawsze nieparzysta dlugosc --> zmiana zwieksza ilosc ciastek o 1
//not really, czasami nie da sie dolaczyc hshs
//claim: przedzialy parzyste na prefiksie i sukfiksie sa useless
//skoro wybieramy a[1], to a[1] >= a[2], dalej jakby tych dwoch elementow w ogole nie bylo
//wtedy wartosci nie maja znaczenia, bo ostatni na prefiksie bedzie pusty
int ans = 0;
auto valid = [&](T x){
auto [val, l] = x;
return val == cost[l] && beg[l] && (R[l]-l)%2 == 0;
};
for (int rep = 1; rep <= (n+1)/2; rep++){
while (!valid(pq.top())) pq.pop();
auto [val, l] = pq.top(); pq.pop();
ans += val;
debug(val, l, R[l]);
int nl = L[l-1];
int nr = R[R[l]+1];
int newcost = cost[nl] + cost[R[l]+1] - cost[l];
nl = max(nl, 1ll);
nr = min(nr, n);
cost[nl] = newcost;
beg[R[l]+1] = beg[l] = 0;
beg[nl] = 1;
pq.push({cost[nl], nl});
R[nl] = nr; L[nr] = nl;
cout << ans << "\n";
}
}
int32_t main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
