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 int;

int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    vector<ll> a(n);
    for(auto &e: a) cin >> e;
    if(n%2 == 0) {
        vector<ll> l(n, -1e15), r(n, -1e15);
        l[0] = a[0];
        l[1] = a[1];
        for(ll i=2; i<n; i++)
            l[i] = a[i] + l[i-2];
        r[n-1] = a[n-1];
        r[n-2] = a[n-2];
        for(ll i=n-3; i>=0; i--)
            r[i] = a[i] + r[i+2];
        ll res = -1e15;
        for(ll i=0; i<n; i++)
            res = max(res, l[i] + r[i+3]);

        cout << max({res, l[n-1], l[n-2]}) << endl;
        return 0;
    }

    vector<ll> l(n, -1e15), r(n, -1e15);
    l[0] = a[0];
    l[1] = a[1];
    for(ll i=2; i<n; i++)
        l[i] = max(a[i] + l[i-2], i-3>=0&&(i&1) ? a[i] + l[i-3] : (ll)-1e15);
    r[n-1] = a[n-1];
    r[n-2] = a[n-2];
    for(ll i=n-3; i>=0; i--)
        r[i] = max(a[i] + r[i+2], i+3<n&&(i&1) ? a[i] + r[i+3] : (ll)-1e15);

    ll res = max({l[n-3], r[2], l[n-2], r[1]});
    for(ll i=0; i<n; i++) {
        if(i%2==0) {
            if(i+3<n) res = max(res, l[i] + r[i+3]);
            if(i+4<n) res = max(res, l[i] + r[i+4]);
        } else {
            if(i+2<n) res = max(res, l[i] + r[i+2]);
            if(i+3<n) res = max(res, l[i] + r[i+3]);
        }
    }
    cout << res << endl;
}