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 <iostream>
#include <vector>
#include <algorithm>

int main()
{
    using ll = long long;
    int N;
    std::cin >> N;

    std::vector<ll> A(N);

    for (int i = 0; i < N; ++i) {
        std::cin >> A[i];
    }

    ll dp[200000][3] = {};
    dp[0][0] = A[0];
    dp[1][1] = A[1];
    dp[2][2] = A[2];

    for (int i = 3; i < N; ++i) {
        dp[i][0] = dp[i - 2][0] + A[i];
        dp[i][1] = std::max(dp[i - 2][1] + A[i], dp[i - 3][0] + A[i]);
        dp[i][2] = std::max(dp[i - 2][2] + A[i], dp[i - 3][1] + A[i]);
    }


    if (N % 2 == 0) {
        std::cout << dp[N - 1][1] << std::endl;
    } else {
        std::cout << dp[N - 1][2] << std::endl;
    }

    return 0;
}