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>
using namespace std;

int main() {
    int64_t N;
    cin >> N;
    vector<int64_t> A(N);
    for (int64_t i = 0; i < N; i++) {
        cin >> A[i];
    }

    //iまでで選べる個数は最小(N / 2) - (N - i) / 2個、最大(i + 2) / 2個
    //これらの差は0か1
    enum State {
        MIN, MAX, STATE_NUM
    };

    //また、直前で選んだかどうかで更に2倍の状態
    //dp[i][j][k] := iまでで、選んだ数がj = MIN or MAXであり、k = (直前で選んだ)場合の最大値
    vector<vector<vector<int64_t>>> dp(N + 1, vector<vector<int64_t>>(STATE_NUM, vector<int64_t>(2, INT_MIN)));
    dp[0][0][0] = 0;

    for (int64_t i = 0; i < N; i++) {
        //選んでいる数として可能な最小値、最大値を計算
        const array<int64_t, 2> curr_num = { max(N / 2 - (N - i + 1) / 2, (int64_t)0), (i + 1) / 2 };
        const array<int64_t, 2> next_num = { N / 2 - (N - i) / 2, (i + 2) / 2 };

        //状態が2×2通り
        for (int64_t j : { MIN, MAX }) {
            for (int64_t k : { 0, 1 }) {
                //遷移が2通り
                for (int64_t add : { 0, 1 }) {
                    if (k == 1 && add == 1) {
                        //連続して選ぶのは不可
                        continue;
                    }
                    const int64_t num = curr_num[j] + add;

                    //遷移先が2通り
                    for (int64_t next_state : { MIN, MAX }) {
                        if (num == next_num[next_state]) {
                            dp[i + 1][next_state][add] = max(dp[i + 1][next_state][add], dp[i][j][k] + A[i] * add);
                        }
                    }
                }
            }
        }
    }

    cout << max({ dp[N][0][0], dp[N][0][1], dp[N][1][0], dp[N][1][1] }) << endl;
}