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;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;

#define REP(i, n, m) for(ll i=n; i<(ll)m; ++i)
#define IREP(i, n, m) for(ll i=n-1; i>=m; --i)
#define rep(i, n) REP(i, 0, n)
#define irep(i, n) IREP(i, n, 0)
#define all(v) v.begin(), v.end()

int main(){
    cout << setprecision(20);

    ll N;
    cin >> N;

    vll a(N, 0);
    rep(i, N) cin >> a[i];

    vvll dp(N, vll(3, 0));
    irep(i, N){
        dp[i][0] = a[i] + ((i+2)<N? dp[i+2][0] : 0);
        if(i+3<N) dp[i][1] = a[i] + ((i+3)<N? max(dp[i+3][0], dp[i+2][1]) : (i+2)<N? dp[i+2][1] : 0);
        if(i+4<N) dp[i][2] = a[i] + ((i+4)<N? max(dp[i+4][0], max(dp[i+3][1], dp[i+2][2])) : (i+3)<N? max(dp[i+3][1], dp[i+2][2]) : (i+2)<N? dp[i+2][2] : 0);
    }
    ll total = N%2==0? max(dp[0][1], dp[1][0]) : max(dp[0][2], max(N>3? dp[1][1] : 0, dp[2][0]));
    cout << total << endl;
}