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)

import sys
from bisect import *
from heapq import *
from collections import *
from itertools import *
from functools import *
from math import *
 
sys.setrecursionlimit(100000000)
input = lambda: sys.stdin.readline().rstrip()

N = int(input())
A = list(map(int, input().split()))

dp = [{} for _ in range(N + 1)]
dp[0] = {0: 0}
dp[1] = {0: 0, 1: A[0]}

for i in range(2, N + 1):
    for j in range(max(0, i // 2 - 2), min(N, i // 2 + 2) + 1):
        candidates = []
        if j - 1 in dp[i - 2]:
            candidates.append(dp[i - 2][j - 1] + A[i - 1])
        if j in dp[i - 1]:
            candidates.append(dp[i - 1][j])
        if len(candidates) > 0:
            dp[i][j] = max(candidates)

print(dp[N][N // 2])