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 <stdio.h>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <numeric>
#include <complex>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <cassert>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
const double EPS = 1e-9;
typedef vector<int> vint;
typedef vector<vector<int>> v2int;
typedef vector<ll> vll;
typedef vector<vector<ll>> v2ll;
typedef list<int> liint;
typedef pair<int, int> pint;
const int INF = int(2e9);
const ll LINF = ll(2e9) * ll(2e9);
#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for(int i = x; i < n; i++)
template<class T, class C> void chmax(T& a, C b) { a > b ? : a = b; }
template<class T, class C> void chmin(T& a, C b) { a < b ? : a = b; }

ll score(int N ,int i, int j, vll& Seven, vll& Sodd) {
    if(i - 1 < 0) return Sodd[N] - Sodd[j] + Seven[j - 1] - Seven[i];
    return Sodd[i - 1] + Sodd[N] - Sodd[j] + Seven[j - 1] - Seven[i];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int N;
    cin >> N;
    vll A(N);
    vll Seven(N  + 1);
    vll Sodd(N  + 1);
    rep(i, N) cin >> A[i];
    rep(i, N) {
        if ((i + 1) % 2 == 0) {
            Seven[i + 1] = Seven[i] + A[i];
            Sodd[i + 1] = Sodd[i];
        }
        else {
            Sodd[i + 1] = Sodd[i] + A[i];
            Seven[i + 1] = Seven[i];
        }
    }
    if (N % 2 == 0) {
        cout << max(Sodd[N], Seven[N]) << endl;
        return 0;
    }
    ll ans = -LINF;
    REP(i, 0, N + 1) {
        REP(j, i + 1, N + 1) {
            if (i % 2 != 0 || j % 2 != 1) continue;
            ans = max(ans, score(N, i, j, Seven, Sodd));
        }
    }
    cout << ans << endl;
    return 0;
}