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 -O #-}

import Data.List (scanl, unfoldr)
import qualified Data.ByteString.Char8 as B
import qualified Data.IntSet as S

import Control.Monad
import Control.Monad.ST
import Data.STRef
import qualified Data.Vector.Unboxed as V
import qualified Data.Vector.Unboxed.Mutable as MV

msort :: MV.STVector s Int -> (Int,Int) -> STRef s Int -> ST s ()
msort xv (s,e) ra
  | n == 1 = return ()
  | True = do
    msort xv lr ra
    msort xv rr ra
    tv <- V.freeze xv
    rc <- newSTRef 0
    msort2 xv tv s lr rr ra rc
  where
    n = e - s
    m = s + n `div` 2
    lr@(ls,le) = (s,m)
    rr@(rs,re) = (m,e)
--
msort2 :: MV.STVector s Int -> (V.Vector Int) -> Int -> (Int,Int) -> (Int,Int) -> STRef s Int -> STRef s Int -> ST s ()
msort2 xv tv s (ls,le) (rs,re) ra rc
  | ls==le && rs==re = return ()
  | rs==re = forM_ [ls..le-1] $ \i -> (MV.write xv (s+i-ls) (tv V.! i) >> liftM2 (+) (readSTRef ra) (readSTRef rc) >>= writeSTRef ra)
  | ls==le = forM_ [rs..re-1] $ \i -> (MV.write xv (s+i-rs) (tv V.! i) >> liftM (+1) (readSTRef rc) >>= writeSTRef rc)
  | l <= r = liftM2 (+) (readSTRef ra) (readSTRef rc) >>= writeSTRef ra >> MV.write xv s l >> msort2 xv tv (s+1) (ls+1,le) (rs,re) ra rc
  | True   = liftM (+1) (readSTRef rc) >>= writeSTRef rc >> MV.write xv s r >> msort2 xv tv (s+1) (ls,le) (rs+1,re) ra rc
    where
      l = tv V.! ls
      r = tv V.! rs

invc ss = runST $ do
  v <- V.thaw $ V.fromList ss :: ST s (MV.STVector s Int)
  ra <- newSTRef 0
  msort v (0, MV.length v) ra
  readSTRef ra

si as x = invc ss
  where
    ss = tail $ scanl (\ac a->if a>=x then ac+1 else ac-1) 0 as

bsearch f (a,b)
  | a+1 == b = a
  | f m  = bsearch f (m,b)
  | True = bsearch f (a,m)
  where m = (a + (b-a)`div`2)

ff av as tn x = iv >= (tn+1) `div` 2
  where
    iv = (-) tn $ si as $ av V.! x

main=do
  n<-readLn::IO Int
  as<-unfoldr (B.readInt.B.dropWhile(==' '))<$>B.getLine::IO[Int]
  av<-pure $ V.fromList $ S.toList $ S.fromList as
  print $ (V.!) av $ bsearch (ff av as $ (n*(n-1))`div`2) (-1,V.length av)