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 <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <iomanip>
#include <stack>
#include <queue>
#include <numeric>
#include <map>
#include <unordered_map>
#include <set>
#include <fstream>
#include <chrono>
#include <random>
#include <bitset>
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define mod 1000000007
using ll = long long;
using namespace std;
ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}

int main(){
  int N; cin >> N;
  vector<ll> A(N); rep(i,N) cin >> A[i];
  ll ans = 0;
  const ll mini = -1e17;
  if(N%2==0){
    vector<ll> dp00(N+1, 0), dp01(N+1, 0), dp10(N+1, 0), dp11(N+1, 0);
    rep(i,N){
      dp00[i+1] = i%2? dp01[i] : -1e17;
      dp01[i+1] = i%2? -1e17: dp00[i]+A[i];
      dp10[i+1] = max(dp00[i], dp11[i]);
      dp11[i+1] = i>0? dp10[i]+A[i]: -1e17;
    }

    ans = max(dp10[N], dp11[N]);
  }
  else{
    vector<ll> dp00(N+1, 0), dp01(N+1, 0), dp10(N+1, 0), dp11(N+1, 0), dp20(N+1, 0), dp21(N+1, 0);

    rep(i,N){
      dp00[i+1] = i%2? dp01[i] : mini;
      dp01[i+1] = i%2? mini: dp00[i] + A[i];
      dp10[i+1] = max(dp00[i], dp11[i]);
      dp11[i+1] = i>0? dp10[i]+A[i]: mini;
      dp20[i+1] = max(dp10[i], dp21[i]);
      dp21[i+1] = i>1? dp20[i]+A[i]: mini;
    }
    /*
    rep(i,N+1) cout << dp00[i] << " ";
    cout << endl;
    rep(i,N+1) cout << dp01[i] << " ";
    cout << endl;
    rep(i,N+1) cout << dp10[i] << " ";
    cout << endl;
    rep(i,N+1) cout << dp11[i] << " ";
    cout << endl;
    rep(i,N+1) cout << dp20[i] << " ";
    cout << endl;
    rep(i,N+1) cout << dp21[i] << " ";
    cout << endl;
    */

    ans = max({dp21[N], dp10[N]});
  }
  cout << ans << endl;
  return 0;
}