Problem:
Score : 600 points

Problem Statement

There is an infinitely large pond, which we consider as a number line.
In this pond, there are N lotuses floating at coordinates 0, 1, 2, ..., N-2 and N-1.
On the lotus at coordinate i, an integer s_i is written.

You are standing on the lotus at coordinate 0. You will play a game that proceeds as follows:

1. Choose positive integers A and B. Your score is initially 0.

2. Let x be your current coordinate, and y = x+A. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

3. Let x be your current coordinate, and y = x-B. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

4. Go back to step 2.

You want to end the game with as high a score as possible.
What is the score obtained by the optimal choice of A and B?

Constraints

3 \leq N \leq 10^5

-10^9 \leq s_i \leq 10^9

s_0=s_{N-1}=0

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
s_0 s_1 ...... s_{N-1}

Output

Print the score obtained by the optimal choice of A and B.

Sample Input 1

5
0 2 5 1 0

Sample Output 1

3

If you choose A = 3 and B = 2, the game proceeds as follows:

Move to coordinate 0 + 3 = 3. Your score increases by s_3 = 1.

Move to coordinate 3 - 2 = 1. Your score increases by s_1 = 2.

Move to coordinate 1 + 3 = 4. The game ends with a score of 3.

There is no way to end the game with a score of 4 or higher, so the answer is 3. Note that you cannot land the lotus at coordinate 2 without drowning later.

Sample Input 2

6
0 10 -7 -4 -13 0

Sample Output 2

0

The optimal strategy here is to land the final lotus immediately by choosing A = 5 (the value of B does not matter).

Sample Input 3

11
0 -4 0 -99 31 14 -15 -39 43 18 0

Sample Output 3

59

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE BangPatterns          #-}
{-# LANGUAGE CPP                   #-}
{-# LANGUAGE LambdaCase            #-}
{-# LANGUAGE MagicHash             #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE MultiWayIf            #-}
{-# LANGUAGE OverloadedStrings     #-}
{-# LANGUAGE TupleSections         #-}
{-# LANGUAGE TypeFamilies          #-}
{-# LANGUAGE ViewPatterns          #-}

import           Control.Applicative
import           Control.Exception
import           Control.Monad
import           Control.Monad.Primitive
import           Control.Monad.ST
import           Control.Monad.State.Strict
import           Data.Bool
import qualified Data.ByteString             as B
import qualified Data.ByteString.Char8       as C
import qualified Data.ByteString.Unsafe      as B
import           Data.Char
import qualified Data.Foldable               as F
import           Data.Function
import qualified Data.IntMap.Strict          as IM
import qualified Data.IntSet                 as IS
import qualified Data.List                   as L
import qualified Data.Map.Strict             as M
import           Data.Monoid
import           Data.Ord
import           Data.Primitive.MutVar
import qualified Data.Set                    as S
import           Data.Tuple
import qualified Data.Vector                 as V
import qualified Data.Vector.Generic         as G
import qualified Data.Vector.Generic.Mutable as GM
import qualified Data.Vector.Unboxed         as U
import qualified Data.Vector.Unboxed.Mutable as UM
import           Debug.Trace
import           Foreign                     hiding (void)
import           GHC.Exts
import           Unsafe.Coerce

main :: IO ()
main = do
    !n <- readLn :: IO Int
    xs <- U.unfoldrN n (C.readInt.C.dropWhile isSpace) <$> C.getLine
    print $ solve n xs

solve :: Int -> U.Vector Int -> Int
solve n xs = F.foldl' max 0 $ map step [1..n - 2]
  where
    step diff
        | rem (n - 1) diff == 0 = go0 0 0 (n - 1) 0
        | otherwise = go 0 0 (n - 1) 0
      where
        go !res !acc !l !r
            -- | traceShow (diff, l, r, acc) False = undefined
            | 0 <= l', r' < n = go (max res acc') acc' l' r'
            | otherwise = res
          where
            l' = l - diff
            r' = r + diff
            acc' = acc + U.unsafeIndex xs l' + U.unsafeIndex xs r'

        go0 !res !acc !l !r
            -- | traceShow (diff, l, r, acc) False = undefined
            | r' <= l' = res
            | 0 <= l', r' < n = go0 (max res acc') acc' l' r'
            | otherwise = res
          where
            l' = l - diff
            r' = r + diff
            acc' = acc + U.unsafeIndex xs l' + U.unsafeIndex xs r'

-------------------------------------------------------------------------------
type Parser a = C.ByteString -> Maybe (a, C.ByteString)

parseInt :: Parser Int
parseInt = C.readInt . C.dropWhile isSpace

parseInt2 :: Parser (Int, Int)
parseInt2 = runStateT $
    (,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt3 :: Parser (Int, Int, Int)
parseInt3 = runStateT $
    (,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt4 :: Parser (Int, Int, Int, Int)
parseInt4 = runStateT $
    (,,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)