Problem:
Score : 600 points

Problem Statement

Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

2 \leq N \leq 2\times 10^5

|A_i|\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the maximum possible sum of the chosen elements.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4

295

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
// Begin Header {{{
#define all(x) (x).begin(), (x).end()
#define rep(i, s, n) for (i64 i = (s), i##_limit = (n); i < i##_limit; ++i)
#define repr(i, s, t) for (i64 i = (s), i##_limit = (t); i >= i##_limit; --i)
#define var(type, ...) \
    type __VA_ARGS__;  \
    read(__VA_ARGS__);
#ifndef DBG
#define dump(...)
#endif
using namespace std;
using i64 = int_fast64_t;
using pii = pair<i64, i64>;
template <class T, class U>
inline bool chmax(T &a, const U &b) {
    return b > a && (a = b, true);
}
template <class T, class U>
inline bool chmin(T &a, const U &b) {
    return b < a && (a = b, true);
}
constexpr int INF = 0x3f3f3f3f;
constexpr i64 LINF = 0x3f3f3f3f3f3f3f3fLL;

template <class T>
inline vector<T> makeV(const T &initValue, size_t sz) {
    return vector<T>(sz, initValue);
}

template <class T, class... Args>
inline auto makeV(const T &initValue, size_t sz, Args... args) {
    return vector<decltype(makeV<T>(initValue, args...))>(
        sz, makeV<T>(initValue, args...));
}

template <class T>
inline istream &operator>>(istream &is, vector<T> &vec) {
    for (auto &e : vec) is >> e;
    return is;
}

inline void read() {}

template <class Head, class... Tail>
inline void read(Head &head, Tail &... tail) {
    cin >> head;
    read(tail...);
}

inline void print() { cout << "\n"; }

template <class Head, class... Tail>
inline void print(Head &&head, Tail &&... tail) {
    cout << head;
    if (sizeof...(tail)) cout << ' ';
    print(forward<Tail>(tail)...);
}

template <class T>
inline ostream &operator<<(ostream &os, const vector<T> &vec) {
    static constexpr const char *delim[] = {" ", ""};
    for (const auto &e : vec) os << e << delim[&e == &vec.back()];
    return os;
}

template <class Container>
struct Rev {
    Container &x_;
    inline Rev(Container &x) : x_(x) {}
    inline auto begin() { return rbegin(x_); }
    inline auto end() { return rend(x_); }
};
// }}} End Header

int N, K;
vector<i64> a;

i64 memo[200010][4];

i64 dp(int i, int cnt) {
    if (i >= N) {
        return (cnt <= K) ? 0 : -LINF;
    }
    if (i > N) return -LINF;
    if (cnt > K) return -LINF;

    auto &ret = memo[i][cnt];
    if (~ret) return ret;

    ret = -LINF;
    chmax(ret, dp(i + 2, cnt) + a[i]);
    chmax(ret, dp(i + 3, cnt + 1) + a[i]);
    return ret;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    read(N);
    a.resize(N);
    read(a);
    K = (N & 1) ? 2 : 1;

    memset(memo, -1, sizeof(memo));

    i64 ans = -LINF;
    chmax(ans, dp(0, 0));
    chmax(ans, dp(1, 1));
    chmax(ans, dp(2, 2));

    print(ans);

    return 0;
}