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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
long long odd_max(long long k, vector<long long> &a);
long long even_max(long long k, vector<long long> &a);



int main() {
  long n;
  cin >> n;
  vector<long long> a(n);
  for (int i=0; i<n; i++){
    cin >> a.at(i);
  }

  if (n%2) cout <<  odd_max(n,a) << endl;
  else cout << even_max(n,a) << endl;
}


long long odd_max(long long k, vector<long long> &a){
  if(k == 1)
    return 0;

  return max(even_max(k-1,a), odd_max(k-2,a) + a.at(k-1));
}

long long even_max(long long k, vector<long long> &a){
  if(k == 2)
    return max(a.at(0),a.at(1));
  
  long long s = 0;
  for(int i=0; i<k;i+=2)
    s += a.at(i);
  
  return max(even_max(k-2,a)+a.at(k-1), s);
}