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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9+7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))

n = I()
a = LI()
if n == 3:
    print(max(a))
    quit()
dp1 = [-inf]*(n+1) 
dp2 = [-inf]*(n+1)
dp3 = [-inf]*(n+1)
dp1[0] = 0
dp2[0] = 0
dp3[0] = 0
dp1[1] = a[0]
dp2[2] = a[1]
for i in range(3,n+1):
    dp1[i] = dp1[i-2] + a[i-1]
    dp2[i] = max(dp2[i-2],dp1[i-3]) + a[i-1]
    dp3[i] = max(dp3[i-2],dp2[i-3],dp1[i-4]) + a[i-1]
if n % 2:
    print(max(dp1[n-2], dp2[n], dp3[n]))
else:
    print(max(dp1[n],dp2[n]))