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;

#define ll long long
#define mx 200005
int n,m;
int ar[mx];
ll dp[mx],dp1[mx];
bool visit[mx][3];



int main()
{
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d",&ar[i]);
    }
    ll re;
    if(n%2==0)
    {
        re=-1e18;
        for(int i=0; i<n/2; i++)
        {
            dp[i+1]=dp[i]+ar[i*2];
            dp1[i+1]=dp1[i]+ar[i*2+1];
        }
        for(int i=1; i<=n/2; i++)
        {
            re=max(re,dp[i]+(dp1[n/2]-dp1[i]));
        }
    }
    else
    {
        ll sum=ar[0];
        for(int i=1; i<=n/2; i++)
        {
            dp[i]=ar[i*2]+max(dp[i-1],dp1[i-1]);
            dp1[i]=max(sum,ar[i*2-1]+dp1[i-1]);
            sum+=ar[2*i];
        }
        re=max(dp[n/2],dp1[n/2]);
    }
    printf("%lld\n",re);
    return 0;

}