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 <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <limits>
#include <queue>
#include <iomanip>

using namespace std;
using ll = long long int;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
const int MOD=1000000007;
//const int MOD=998244353;
const int INTMAX=1001001000;
const ll LLMAX=1010010010010010000;


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    ll N;
    cin>>N;
    vll A(N);
    for(int i=0;i<N;i++)cin>>A[i];
    
    if(N%2==0){
        vvll dp(N/2,vll(2,0));//奇数にシフト
        for(int i=0;i<N/2;i++){
            if(i==0){
                dp[0][0]=A[0];
                dp[0][1]=A[1];
            }else{
                dp[i][0]=A[i*2]+dp[i-1][0];
                dp[i][1]=max(dp[i-1][1],dp[i-1][0])+A[2*i+1];
            }
        }
        cout<<max(dp[N/2-1][0],dp[N/2-1][1])<<endl;
    }else{
        vvll dp(N/2,vll(3,0));
        for(int i=0;i<N/2;i++){
            if(i==0){
                dp[0][0]=A[0];
                dp[0][1]=A[1];
                dp[0][2]=A[2];
            }else{
                dp[i][0]=A[i*2]+dp[i-1][0];
                dp[i][1]=max(dp[i-1][1],dp[i-1][0])+A[2*i+1];
                dp[i][2]=max(dp[i-1][2],dp[i-1][1])+A[i*2+2];
            }
        }
        cout<<max(dp[N/2-1][2],max(dp[N/2-1][0],dp[N/2-1][1]))<<endl;
    }
    
    return 0;
}