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)

from itertools import accumulate
n,*a=map(int,open(0).read().split())
if n%2==0:
	o=a[::2]
	e=a[1::2]
	o=[0]+list(accumulate(o))
	e=list(accumulate(e[::-1]))[::-1]+[0]
	ans=0
	for i in range(n//2+1):
		ans=max(ans,o[i]+e[i])
	print(ans)
	exit()
o=a[::2]
e=a[1::2]
o=[0]+list(accumulate(o))
e=list(accumulate(e[::-1]))[::-1]+[0]
dp=[-1e100]*n+[0]
for i in range(n):
	if i%2==1:
		dp[i]=max(o[i//2]+a[i-1],o[i//2]+a[i],dp[i-2]+a[i])
	elif i>0:
		dp[i]=max(dp[i-1],dp[i-2]+a[i])
print(dp[-2])