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 Haskell (GHC 8.8.3) (judge verdict: Runtime Error)

main = do
  li <- getLine
  let n = read li
  li <- getLine
  let as = map read $ words li
  let ans = compute n as
  print ans

{-
偶数個のとき、奇数番目の和と偶数番目の和の大きい方。
奇数個のとき、まず列から最小値を見つけ、それの前と後ろに分ける。
どちらも偶数個なので、上の戦略でそれぞれ大きい値を見つけ、足す。
最小値が複数あるかもしれないのが面倒。それらの間で部分和を共有しないとか？

分割して両方奇数個になる場合もあるので間違い。
-}

compute :: Int -> [Int] -> Int
compute n as
  | even n = computeEven 0 0 as
  | True   = maximum vs
  where
    x = minimum as
    is = map snd $ filter ((x ==).fst) $ zip as [0..]
    vs = [ computeEven 0 0 bs + computeEven 0 0 (drop 1 cs)
         | i <- is, let (bs,cs) = splitAt i as ]

computeEven s t [] = max s t
computeEven s t (x:y:xs) = computeEven (s+x) (t+y) xs