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;

int n;
int max_gap;
vector<long long> a;

long long used[3][200002];
long long dp[3][200002];

long long calc(int gap, int idx) {
    if (idx < 0) return 0;
    if (used[gap][idx]) return dp[gap][idx];
    long long ans = - (1LL << 60);
    for (int i = 0; i <= gap; i++) {
        ans = max(ans, a[idx] + calc(gap - i, idx - 2 - i));
        if (i != 0) ans = max(ans, calc(gap - i, idx - i));
    }
    used[gap][idx] = true;
    return dp[gap][idx] = ans;
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        long long x;
        cin >> x;
        a.push_back(x);
    }
    max_gap = n % 2 == 0 ? 1 : 2;
    
    cout << calc(max_gap, n - 1) << endl;
    return 0;
}