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>
#define ll long long int
using namespace std;
int main(){
 ll n,c=0;
   cin >> n;
  ll arr[n];
  for(ll i=0;i<n;i++){
    cin >> arr[i];
  }
   if(n%2 ==0){
     ll sume=0,sumo=0;
   for(int i=0;i<n;i=i+2){
     sume = sume + arr[i];
   }
     for(int i=1;i<n;i=i+2){
       sumo =sumo + arr[i];
   }
     cout << max(sume,sumo) << "\n";
   }else{
      ll sume=0,sumo=0,sum=0;
   for(int i=0;i<n-2;i=i+2){
     sume = sume + arr[i];
   }
     for(int i=1;i<n-1;i=i+2){
       sumo =sumo + arr[i];
   }
     for(int i=2;i<n;i=i+2){
       sum =sum + arr[i];
     }
     cout << max(sum,max(sume,sumo)) <<"\n";
      }
  return 0;
}