Problem:
Score : 700 points

Problem Statement

There is an integer sequence of length 2^N: A_0, A_1, ..., A_{2^N-1}. (Note that the sequence is 0-indexed.)

For every integer K satisfying 1 \leq K \leq 2^N-1, solve the following problem:

Let i and j be integers. Find the maximum value of A_i + A_j where 0 \leq i < j \leq 2^N-1 and (i or j) \leq K.
Here, or denotes the bitwise OR.

Constraints

1 \leq N \leq 18

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 ... A_{2^N-1}

Output

Print 2^N-1 lines.
In the i-th line, print the answer of the problem above for K=i.

Sample Input 1

2
1 2 3 1

Sample Output 1

3
4
5

For K=1, the only possible pair of i and j is (i,j)=(0,1), so the answer is A_0+A_1=1+2=3.

For K=2, the possible pairs of i and j are (i,j)=(0,1),(0,2).
When (i,j)=(0,2), A_i+A_j=1+3=4. This is the maximum value, so the answer is 4.

For K=3, the possible pairs of i and j are (i,j)=(0,1),(0,2),(0,3),(1,2),(1,3),(2,3) .
When (i,j)=(1,2), A_i+A_j=2+3=5. This is the maximum value, so the answer is 5.

Sample Input 2

3
10 71 84 33 6 47 23 25

Sample Output 2

81
94
155
155
155
155
155

Sample Input 3

4
75 26 45 72 81 47 97 97 2 2 25 82 84 17 56 32

Sample Output 3

101
120
147
156
156
178
194
194
194
194
194
194
194
194
194

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize ("O3")
#include "bits/stdc++.h"

using namespace std;
using ll = long long int;

#define debug(v) {printf("L%d %s > ",__LINE__,#v);cout<<(v)<<endl;}
#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:(v)){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(remove_const<decltype(l)>::type cnt=0;(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define diterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);--(cnt))
const ll MD = 1000000007ll; const long double PI = 3.1415926535897932384626433832795L;
inline void assert_call(bool assertion, function<void()> f) { if (!assertion) { cerr << "assertion fault:" << endl; f(); abort(); } }
template<typename T1, typename T2> inline ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << '(' << p.first << ':' << p.second << ')'; return o; }
template<typename Vec> inline ostream& _ostream_vecprint(ostream& os, const Vec& a) {
    os << '['; for (const auto& e : a) os << ' ' << e << ' '; os << ']'; return os;
}
template<typename T> inline ostream& operator<<(ostream& o, const vector<T>& v) { return _ostream_vecprint(o, v); }
template<typename T, size_t S> inline ostream& operator<<(ostream& o, const array<T, S>& v) { return _ostream_vecprint(o, v); }
template<typename T> inline T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> inline T& minset(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
template<typename T> inline T rand(T l, T h) { return uniform_int_distribution<T>(l, h)(randdev); }
template<> inline double rand<double>(double l, double h) { return uniform_real_distribution<double>(l, h)(randdev); }
template<> inline float rand<float>(float l, float h) { return uniform_real_distribution<float>(l, h)(randdev); }

#if defined(_WIN32) || defined(_WIN64)
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#elif defined(__GNUC__)
#else
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0; T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var * sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }
        inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiblechar(cc); cc = getchar_unlocked());
            for (; isvisiblechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
            return *this;
        }
        template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
    };
    class MaiPrinter {
    public:
        template<typename T>
        void output_integer(T var) {
            if (var == 0) { putchar_unlocked('0'); return; }
            if (var < 0)
                putchar_unlocked('-'),
                var = -var;
            char stack[32]; int stack_p = 0;
            while (var)
                stack[stack_p++] = '0' + (var % 10),
                var /= 10;
            while (stack_p)
                putchar_unlocked(stack[--stack_p]);
        }
        inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }
        inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }
        inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }
        inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }
        inline MaiPrinter& operator<<(const string& str) {
            const char* p = str.c_str();
            const char* l = p + str.size();
            while (p < l) putchar_unlocked(*p++);
            return *this;
        }
        template<typename IT> void join(IT begin, IT end, char sep = ' ') { for (bool b = 0; begin != end; ++begin, b = 1) b ? *this << sep << *begin : *this << *begin; }
    };
}
MaiScanner scanner;
MaiPrinter printer;


typedef ll tsearch_t;
typedef ll fval_t;
tsearch_t tri_search_minimize(tsearch_t left, tsearch_t right, function<fval_t(tsearch_t)> func) {

    for (int lop = 0; lop < 60; ++lop) {
        tsearch_t l = right - (2 * right - 2 * left) / 3;
        tsearch_t r = right - (right - left) / 3;
        fval_t lval = func(l);
        fval_t rval = func(r);

        if (lval <= rval)
            right = r;
        else
            left = l;

        if (right - left <= 2) break;
    }
    fval_t best = func(left);
    tsearch_t bestidx = left;
    for (tsearch_t t = left + 1; t <= right; ++t) {
        tsearch_t f = func(t);
        if (best > f) {
            best = f; bestidx = t;
        }
    }
    return bestidx;
}


ll N;
ll A[1 << 19];

ll Amax[1 << 19];

ll ans[1 << 19];

int main() {

    scanner >> N;
    scanner.in(A, A + (1 << N));

    // Amax[0] = A[0];
    // iterate(i, 1, 1 << N) {
    //     ll p = 1 << 24;
    //     while ((i&p) == 0) p >>= 1;
    //     Amax[i] = max(A[i], Amax[i^p]);
    // }

    // yarudake
    repeat(i, (1 << N) - 1) {
        iterate(j, i + 1, 1 << N) {
            ll k = i | j;
            maxset(ans[k], A[i] + A[j]);
        }
    }

    // iterate(k, 1, 1 << N) {
    //     ll p = 1 << 24;
    //     while ((k&p) == 0) p >>= 1;
    //     maxset(ans[k], max(ans[k^p], A[k]+Amax[k^p]));
    // }

    ll curr = 0;
    iterate(k, 1, 1 << N) {
        maxset(curr, ans[k]);
        printer << curr << '\n';
    }

    return 0;
}