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: Accepted)

from collections import defaultdict
INF = -1000000000000001

n = int(input())
l = list(map(int, input().split()))

a = int(n/2)

dpa = defaultdict(lambda: -INF)
#dpa = [[0 for j in range(a+1)] for i in range(n+1)]

dpa[(0,0)] = 0
dpa[(1,0)] = 0
dpa[(1,1)] = l[0]
j = 1
for i in range(2,n+1):
	if i == j*2:
		dpa[(i,j)] = max(dpa[(i-2,j-1)]+l[i-1],dpa[(i-1,j)])
#		print(i,j,dpa[(i,j)])
	else:
		dpa[(i,j)] = max(dpa[(i-2,j-1)]+l[i-1],dpa[(i-1,j)])
#		print(i,j,dpa[(i,j)])
		if (i != n):
			j += 1
#			print(i,j,dpa[(i,j)])
#			print("   ",i-2,j-1,dpa[(i-2,j-1)],"")
			dpa[(i,j)] = dpa[(i-2,j-1)]+l[i-1]

#print(dpa)
print(dpa[n,a])