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 Python (3.8.2) (judge verdict: Accepted)

from collections import defaultdict, Counter
from heapq import heapify, heappop, heappush
from sys import stdin


def main():
    n = int(input())
    a = [int(x) for x in input().split()]
    if n % 2 == 0:
        if n == 2:
            print(max(a))
            return
        v = [[None for j in range(2)] for i in range(n)]
        v[0][0] = a[0]
        v[1][1] = a[1]
        for i in range(2, n):
            for j in range(2):
                c = []
                if v[i-2][j] is not None:
                    c.append(v[i-2][j])
                if j >= 1 and i >= 3:
                    if v[i-3][j-1] is not None:
                        c.append(v[i-3][j-1])
                if len(c) == 0:
                    continue
                v[i][j] = max(c) + a[i]
        c = [v[n-1][j] for j in range(2) if v[n-1][j] is not None]
        print(max(c))
    else:
        if n == 3:
            print(max(a))
            return
        v = [[None for j in range(3)] for i in range(n)]
        v[0][0] = a[0]
        v[1][1] = a[1]
        v[2][0] = a[0]+a[2]
        v[2][2] = a[2]
        for i in range(3, n):
            for j in range(3):
                c = []
                if v[i-2][j] is not None:
                    c.append(v[i-2][j])
                if j >= 1:
                    if v[i-3][j-1] is not None:
                        c.append(v[i-3][j-1])
                if j >= 2 and i >= 4:
                    if v[i-4][j-2] is not None:
                        c.append(v[i-4][j-2])
                if len(c) == 0:
                    continue
                v[i][j] = max(c) + a[i]
        c = [v[n-1][j] for j in range(1, 3) if v[n-1][j] is not None]
        print(max(c))


def input(): return stdin.readline().rstrip()


main()