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: Wrong Answer)

#include <bits/stdc++.h>  // clang-format off
using Int = long long;
#define REP_(i, a_, b_, a, b, ...) for (Int i = (a), lim##i = (b); i < lim##i; i++)
#define REP(i, ...) REP_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
struct SetupIO { SetupIO() { std::cin.tie(nullptr), std::ios::sync_with_stdio(false), std::cout << std::fixed << std::setprecision(13); } } setup_io;
#ifndef _MY_DEBUG
#define dump(...)
#endif  // clang-format on

/**
 *    author:  knshnb
 *    created: Sun Apr 12 21:00:01 JST 2020
 **/

template <class T, class S> T make_vec(S x) { return x; }
template <class T, class... Ts> auto make_vec(size_t n, Ts... ts) {
    return std::vector<decltype(make_vec<T>(ts...))>(n, make_vec<T>(ts...));
}

signed main() {
    Int n;
    std::cin >> n;
    std::vector<Int> a(n);
    REP(i, n) std::cin >> a[i];
    if (n % 2 == 0) {
        Int even = 0;
        REP(i, n / 2) { even += a[2 * i]; }
        std::cout << std::max(even, std::accumulate(a.begin(), a.end(), -even)) << std::endl;
        return 0;
    }
    auto dp = make_vec<Int>(n + 1, 3, 0);
    REP(i, n) {
        dp[i + 1][0] = dp[i][0] + (i % 2 ? 0 : a[i]);
        dp[i + 1][1] = std::max(dp[i][0], dp[i][1] + (i % 2 ? a[i] : 0));
        if (i % 2) {
            dp[i + 1][2] = std::max({dp[i][1], dp[i][2] + (i % 2 || i == 0 ? 0 : a[i])});
        } else {
            dp[i + 1][2] = std::max({dp[i][1], dp[i][2] + (i % 2 || i == 0 ? 0 : a[i])});
        }
    }
    std::cout << std::max({dp[n - 2][0], dp[n][1], dp[n][2]}) << std::endl;
}