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: Runtime Error)

N = int(input())
A = list(map(int,input().split()))
dp = [[0]*3 for _ in range(N+1)]

dp[1][3] = A[0]
dp[2][2] = max(A[1],dp[1][3])

for i in range(2,N):
  a = A[i]
  mid = (i+1)//2
  ma = (i+2)//2
  mi = i//2
  dp[i+1][2+ma-mid] = max(dp[i][2+ma-i//2],dp[i-1][1+ma-(i-1)//2]+a)
  dp[i+1][2+mi-mid] = max(dp[i][2+mi-i//2],dp[i-1][1+mi-(i-1)//2]+a)

print(dp[-1][2])