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>
#define pii pair<int,int>
#define pb push_back
#define nmax 200011
#define lld long long
#define inf 9999999999999999LL
using namespace std;
int a[nmax];
lld dp[nmax], st[nmax];
int main()
{
    int n;
    cin>>n;
    for (int i=1;i<=n;++i){
        cin>>a[i];
    }
    st[1] = a[1];
    for (int i=2;i<=n;++i){
        st[i] = st[i-2] + a[i];
    }
    for (int i=2;i<=n;++i){
        if (i % 2){
            dp[i] = max(dp[i-1], dp[i-2] + a[i]);
        } else {
            dp[i] = a[i] + dp[i-2];
            dp[i] = max(dp[i], st[i-1]);
        }
    }
    cout<<dp[n]<<'\n';
    return 0;
}