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>
#include <vector>


int main(){
  int64_t N;
  std::cin >> N;
  
  std::vector< int > vec(N,0);
  for(int cnt = 0; cnt < N; cnt++){
    std::cin >> vec.at(cnt);
  }
  
  const int64_t skip = N%2 + 1;
  std::vector< std::vector< int64_t > > dp( N+2 , std::vector< int64_t >( skip+1, 0 ) );
  
  for(int i = 0; i <= N ; i++){
    for(int64_t j = 0; j < skip ; j++){
      if(j < skip){
        if(dp.at(i+1).at(j+1) < dp.at(i).at(j)){
          dp.at(i+1).at(j+1) = dp.at(i).at(j);
        }
      }
      if(i < N){
        if(dp.at(i+2).at(j) < dp.at(i).at(j)+vec.at(i)){
          dp.at(i+2).at(j) = dp.at(i).at(j)+vec.at(i);
        }
      }
    }
  }
  
  int64_t answer = 0;
  for(int i = 0; i < N+2; i++){
    for(int64_t j = 0; j < skip+1; j++){
      if(answer < dp.at(i).at(j)){
        answer = dp.at(i).at(j);
      }
    }
  }
  
  std::cout << answer << std::endl;

  return 0;
}