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;
using ll = long long;
#define rep(i, begin, n) for (int i = begin; i < n; i++)
#define repe(i, begin, n) for (int i = begin; i <= n; i++)
#define repr(i, begin, n) for (int i = begin; i > begin - n; i--)
#define repre(i, begin, end) for (int i = begin; i >= end; i--)

template <class T>
inline bool chmax(T& a, T b)
{
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}

template <class T>
inline bool chmin(T& a, T b)
{
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

const int inf = 1000000007;
const int MOD = 1000000007;
const long long INF = 1000000000000000007;

// -------------------------------------------------------
ll dp[201010][3];
int N;
ll A[201010];

int main()
{
    cin >> N;
    repe(i, 1, N)
    {
        cin >> A[i];
    }

    /*
    if (N % 2 == 0) {
        ll sum1 = 0;
        ll sum2 = 0;
        repe(i, 1, N / 2)
        {
            sum1 += A[2 * i];
            sum2 += A[2 * i - 1];
        }
        cout << max(sum1, sum2) << endl;
        return 0;
    }
    */

    dp[1][2] = A[1];
    dp[2][1] = max(A[1], A[2]);
    dp[3][1] = max(dp[2][1], A[3]);
    dp[3][2] = dp[1][1] + A[3];

    repe(i, 4, N)
    {
        if (i % 2 == 0) {
            for (int j = 0; j <= 1; j++) {
                dp[i][j] = max(dp[i - 1][j + 1], dp[i - 2][j] + A[i]);
            }
        } else {
            for (int j = 0; j <= 2; j++) {
                dp[i][j] = max(dp[i - 1][j], dp[i - 2][j] + A[i]);
            }
        }
    }

    cout << dp[N][1] << endl;
}