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;
const long long MOD = 1e9+7;
//const long long MOD = ;
const long long INF_LL = 1e18+3; const int INF_int = 1e9+3;
#define ll long long
#define rep(i,n) for(int i = 0;i<n;i++)
#define lrep(i,n) for(long long i = 0;i<n;i++)
#define All(a) a.begin(),a.end()
#define Debug_Output_ALL(a) for(auto itr = a.begin();itr!=a.end();itr++) cout << *itr << " ";cout << endl;
int main(){
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    if(n%2==0){
        ll ans1 = 0;
        ll ans2 = 0;
        for(int i = 0;i<n;i+=2) ans1+=a[i];
        for(int i = 1;i<n;i+=2) ans2+=a[i];
        cout << max(ans1,ans2) << endl;
    }    
    else{
        int nn = (n+1)/2;
        ll left[nn+1],right[nn];
        ll left2[nn+1],right2[nn];
        left[0] = right[0] = 0;
        left2[0]=right2[0]=0;
        ll ans = -INF_LL;
        for(int i = 0;i<nn;i++){
            left[i+1]=left[i]+a[2*i];
            if(i!=nn-1)right[i+1]=right[i]+a[n-2-2*i];
        }
        for(int i = 0;i<nn;i++){
            left2[i+1]=left2[i]+a[n-1-2*i];
            if(i!=nn-1)right2[i+1]=right2[i]+a[2*i+1];
        }
        /*rep(i,nn+1) cout << left2[i] << " ";
        cout << endl;
        rep(i,nn+1) cout << right2[i] << " ";
        cout << endl;*/
        for(int i = 0;i<=nn-1;i++){
            ans = max(ans,left[i]+right[(nn-1-i)]);
            ans = max(ans,left2[i]+right2[(nn-1-i)]);
        }
        cout << ans << endl;
    }
}