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: Accepted)

#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 60;
int main() {
    int N;
    cin >> N;
    vector<long long> A(N);
    for(int i = 0; i < N; i++) {
        cin >> A[i];
    }

    vector<long long> dpfl(200010, -INF), dpc(200010, -INF);
    dpfl[N-1] = 0LL; dpc[N-1] = A[N-1];
    if(N >= 2){
        dpfl[N-2] = max(A[N-2], A[N-1]);
        dpc[N-2] = max(A[N-2], A[N-1]);
    }
    if(N >= 3) {
        dpfl[N-3] = max({A[N-3], A[N-2], A[N-1]});
        dpc[N-3] = A[N-3] + A[N-1];
    }

    for (int i = N - 4; i >= 0; i--)
    {
        if((N-i) % 2 == 0) {
            dpfl[i] = dpc[i] = max(A[i] + dpfl[i+2], A[i+1] + dpc[i+3]);
        }
        else {
            dpfl[i] = max({A[i] + dpfl[i+2], A[i+1] + dpfl[i+3], dpc[i+2]});
            dpc[i] = A[i] + dpc[i+2];
        }
    }

    cout << dpfl[0] << endl;
    return 0;
}