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)

// -----------------------------------
// Author     : MatsuTaku
// Country    : Japan
// Created    : 04/12/20 20:48:37
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n; cin>>n;
  vector<ll> A(n); for(auto& a:A) cin>>a;
  if (n%2==0) {
    ll osum = 0, esum = 0;
    for (int i = 0; i < n; i++) {
      if (i%2==0)
        osum+=A[i];
      else
        esum+=A[i];
    }
    cout << max(osum,esum) << endl;
  } else {
    vector<array<ll, 3>> dp(n);
    dp[0][0] = A[0];
    dp[1][1] = A[1];
    dp[2][2] = A[2];
    dp[2][0] = A[0]+A[2];
    for (int i = 3; i < n; i++) {
      dp[i][0] = dp[i-2][0]+A[i];
      dp[i][1] = max(dp[i-2][1], dp[i-3][0])+A[i];
      dp[i][2] = max(dp[i-2][2], dp[i-3][1]);
      if (i>3)
        dp[i][2] = max(dp[i][2], dp[i-4][0]);
      dp[i][2] += A[i];
    }
    ll ans = max({dp[n-3][0], dp[n-2][1], dp[n-1][2]});
    cout << ans << endl;
  }

  return 0;
}