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 int long long

int n,arr[200005],ps[200005];

void main2(){
  int sum1=0,sum2=0;
  for(int x=0;x<n;x++){
    if((x%2)==0)sum2+=arr[x];
    else sum1+=arr[x];
  }
  cout<<max(sum1,sum2)<<'\n';
}

int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(0);
    int sum=0,ans=LLONG_MIN,maxi=LLONG_MIN;
    cin>>n;
    for(int x=0;x<n;x++){
      cin>>arr[x];
    }
    if((n%2)==0){
      main2();
      return 0;
    }
    for(int x=0;x<n;x++){
      if((x%2)==0){//odd elements...
        sum+=arr[x];
        arr[x-1]-=arr[x];
      }
    }
    for(int x=0;x<n;x++){
      ps[x+1]=ps[x];
      if((x%2)==1){//even elements...
        ps[x+1]+=arr[x];
      }
    }
    maxi=ps[n];
    ans=sum-arr[n-1];
    for(int x=n-1;x>=1;x-=2){
      maxi=max(maxi,ps[x-1]);
      ans=max(ans,sum+maxi-ps[x-1]-arr[x-2]);
    }
    cout<<ans<<'\n';
    return 0;
}
/*

*/