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<iostream>
using namespace std;
long long int A[200005],zero[200005],one[200005];
int main(){
  int N;
  cin>>N;
  for(int i=0; i<N; i++){
    cin>>A[i];
  }
  for(int i=0; i<N; i+=2){
    zero[i]=zero[i+1]=zero[i-1]+A[i];
  }
  for(int i=1; i<N; i+=2){
    one[i]=one[i+1]=one[i-1]+A[i];
  }
  if(N%2==0){
    cout<<max(zero[N-1],one[N-1])<<endl;
    return 0;
  }
  long long int res=max(zero[N-1]-zero[0],one[N-1]);
  res=max(res,max(zero[N-2],one[N-2]));
  res=max(res,max(one[N-1],zero[N-1]));
  for(int i=1; i<N-1; i++){
    res=max(res,zero[i-1]+one[N-1]-one[i]);
    res=max(res,one[i-1]+zero[N-1]-zero[i]);
  }
  cout<<res<<endl;
  return 0;
}