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;

using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  vector<ll> a(n);
  vector<pair<ll, ll>> b(n);
  for (int i = 0; i < n; i++) {
    cin >> a.at(i);
    b.at(i).first = a.at(i);
    b.at(i).second = i;
  }
  
  sort(b.rbegin(), b.rend());
  
  vector<bool> used(n, false);
  ll ans = 0;
  if (n % 2 == 1) {
    ll count = 0;
    for (int i = 0; i < n; i++) {
      if (b.at(i).second == 0) {
        if (used.at(b.at(i).second + 1) == false) {
          used.at(b.at(i).second) = true;
          ans += b.at(i).first;
          count++;
        }
      }
      else if (b.at(i).second == n - 1) {
        if (used.at(b.at(i).second - 1) == false) {
          used.at(b.at(i).second) = true;
          ans += b.at(i).first;
          count++;
        }
      }
      else if (used.at(b.at(i).second - 1) == false && used.at(b.at(i).second + 1) == false) {
        used.at(b.at(i).second) = true;
        ans += b.at(i).first;
        count++;
      }
      if (count >= n / 2) break;
    }
  }
  else {
    ll keep = 0;
    for (int i = 0; i < n; i += 2) keep += a.at(i);
    ans = keep;
    for (int i = n - 2; i >= 0; i -= 2) {
      keep -= a.at(i);
      keep += a.at(i + 1);
      ans = max(ans, keep);
    }
  }
  
  cout << ans << '\n';
}