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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
long long a[200005],sum[200005],f[200005];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        sum[i]=(i>=2?sum[i-2]:0)+a[i];
    }
    for(int i=2;i<=n;i++)
    {
        if(i&1)
        {
            f[i]=max(f[i-2]+a[i],f[i-1]);
        }
        else f[i]=max(f[i-2]+a[i],sum[i-1]);
    }
    cout<<f[n];
}