Problem:
Score : 700 points

Problem Statement

We will define the median of a sequence b of length M, as follows:

Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.

For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N.
For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a.
We will list m_{l, r} for all pairs (l, r) to create a new sequence m.
Find the median of m.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the median of m.

Sample Input 1

3
10 30 20

Sample Output 1

30

The median of each contiguous subsequence of a is as follows:

The median of (10) is 10.

The median of (30) is 30.

The median of (20) is 20.

The median of (10, 30) is 30.

The median of (30, 20) is 30.

The median of (10, 30, 20) is 20.

Thus, m = (10, 30, 20, 30, 30, 20) and the median of m is 30.

Sample Input 2

1
10

Sample Output 2

10

Sample Input 3

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3

8

Solution in Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE TupleSections #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE BangPatterns #-}

import Control.Monad
import Control.Monad.ST
import Control.Applicative
import Control.Monad.Primitive
import Data.List
import Data.Maybe
import Data.Bits
import qualified Data.ByteString.Char8 as BS
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.IntMap as IM

import Data.Time


readIntBS = fst . fromJust . BS.readInt

main :: IO ()
main = do
  n <- readLn :: IO Int
  as <- map readIntBS . BS.words <$> BS.getLine :: IO [Int]
  fen <- newFenwickTree (2*n+1) :: IO (IOUFenwickTree Int)

  let xs = VU.fromList $ sort as

  let
    m = n*(n+1)`div`2
    solve l r
      | l == r = return l
      | l+1 == r = aux (xs VU.! r) >>= \a -> return $ if a then r else l
      | otherwise = do
          a <- aux $ (xs VU.! ((l+r)`div`2))
          if a then
            solve ((l+r)`div`2) r
            else
            solve l ((l+r)`div`2)
        where
          aux x = do
            let
              as' = map (+(n+1)) . scanl (+) 0 $ map (\a -> if a >= x then 1 else -1) as :: [Int]

            VUM.set fen 0
            y <- sum <$> (forM as' $ \a -> queryFenwickTree fen a >>= \q -> modifyFenwickTree fen a >> return q)
            return $ y >= m`div`2
            

  
  --x <- getCurrentTime
  print =<< (xs VU.!) <$> solve 0 (n-1)
  --y <- getCurrentTime
  --print $ diffUTCTime y x


type IOUFenwickTree a = VUM.IOVector a

{-# INLINE newFenwickTree #-} 
newFenwickTree :: Int -> IO (IOUFenwickTree Int)
newFenwickTree n = VUM.replicate (n+1) 0


{-# INLINE modifyFenwickTree #-}
modifyFenwickTree :: IOUFenwickTree Int -> Int -> IO ()
modifyFenwickTree tree i = do
  aux i
  where
    size = VUM.length tree
    {-# INLINE aux #-}
    aux j = do
      when (j <= size) $ do
        VUM.unsafeWrite tree j =<< (+1) <$> VUM.unsafeRead tree j
        aux (j + (j .&. (-j)))

{-# INLINE queryFenwickTree #-}
queryFenwickTree :: IOUFenwickTree Int -> Int -> IO Int
queryFenwickTree tree i = do
  aux i
  where
    size = VUM.length tree
    {-# INLINE aux #-}
    aux j
      | j <= 0 = return 0
      | otherwise = (+) <$> VUM.unsafeRead tree j <*> aux (j - (j .&. (-j)))