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: Accepted)

#include<bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<ll,ll> P;

#define INF -10000000000000

int main(){
    ll d[3][200000];
    ll a[200000];
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        d[0][i]=d[1][i]=d[2][i]=INF;
        cin>>a[i];
    }
    d[0][0]=0;
    d[1][1]=0;
    d[2][2]=0;
    for(int i=2;i==2;i++){
        d[0][i]=d[0][i-2]+a[i-2];
    }
    for(int i=3;i==3;i++){
        d[0][i]=d[0][i-2]+a[i-2];
        d[1][i]=max(d[1][i-2]+a[i-2],d[0][i-3]+a[i-3]);
    }
    for(int i=4;i<n;i++){
        d[0][i]=d[0][i-2]+a[i-2];
        d[1][i]=max(d[1][i-2]+a[i-2],d[0][i-3]+a[i-3]);
        d[2][i]=max(d[2][i-2]+a[i-2],max(d[1][i-3]+a[i-3],d[0][i-4]+a[i-4]));
        //cout<<i<<":"<<d[0][i]<<":"<<d[1][i]<<":"<<d[2][i]<<endl;
    }
    ll sum=0;
    if((n+1)%2){
        sum=max(d[0][n-2]+a[n-2],d[1][n-1]+a[n-1]);
    }else{
        sum=max(d[0][n-3]+a[n-3],max(d[1][n-2]+a[n-2],d[2][n-1]+a[n-1]));
    }
    cout<<sum;

    return 0;
}