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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <typeinfo>
#include <numeric>
#include <functional>
#include <unordered_map>
#include <bitset>
#include <stack>
#include <assert.h>
#include <unordered_set>
#include <random>


using namespace std;
using ll = long long;
using ull = unsigned long long;

const ll INF = 1e18;
const ll MOD = 1e9 + 7;

#define REP(i, n) for(ll i = 0; i < n; i++)

















int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    REP(i, n){
        cin >> a[i];
    }
    if(n % 2 == 1){
        ll s1 = 0, s2 = 0, mn = INF;
        REP(i, n){
            if(i % 2 == 0){
                s1 += a[i];
                mn = min(mn, a[i]);
            }
            else{
                s2 += a[i];
            }
        }
        ll ans = max(s1 - mn, s2);
        vector<ll> r1 = a, r2 = a;
        REP(i, n - 2){
            if(i % 2 == 0){
                r1[i + 2] += r1[i];
            }
            else{
                r2[i + 2] += r2[i];
            }
        }
        ans = max({ans, r1[n - 1] - r1[0], r2[n - 2] ,r1[0] + r2[n - 2] - r2[1]});
        for(ll i = 2; i < n; i++){
            if(i % 2 == 0){
                ans = max(ans, r1[i - 2] + r1[n - 1] - r1[i]);
                ans = max(ans, r1[i - 2] + r2[n - 2] - r2[i - 1]);
                ans = max(ans, r2[i - 1] + r1[n - 1] - r1[i]);
                ans = max(ans, r2[i - 1] + r2[n - 2] - r2[i - 1]);
            }
            else{
                ans = max(ans, r1[i - 1] + r2[n - 2] - r2[i]);
                ans = max(ans, r2[i - 2] + r1[n - 1] - r1[i - 1]);
            }
        }
        ans = max(ans, r1[n - 1] - r1[0]);
        ans = max(ans, r1[0] + r1[n - 1] - r1[2]);
        ans = max(ans, r1[0] + r2[n - 2] - r2[1]);
        for(ll i = 2; i < n - 1; i++){
            if(i % 2 == 0){
                ans = max(ans, r1[n - 1] - r1[i] + r1[i - 2]);
                ans = max(ans, r1[n - 1] - r1[i] + r2[i - 1]);
            }
            else{
                ans = max(ans, r1[i - 1] + r1[n - 1] - r1[i + 1]);
                ans = max(ans, r1[i - 1] + r2[n - 2] - r2[i]);
            }
        }
        cout << ans << endl;
        return 0;
    }
    vector<ll> r1 = a, r2 = a;
    REP(i, n - 2){
        if(i % 2 == 0){
            r1[i + 2] += r1[i];
        }
        else{
            r2[i + 2] += r2[i];
        }
    }
    ll ans = max(r1[n - 2], r2[n - 1]);
    for(ll i = n - 2; i > 0; i -= 2){
        ans = max(ans, r1[i - 2] + r2[n - 1] - r2[i - 1]);
    }
    cout << ans << endl;
}