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)

import Control.Applicative
main = do 
    getLine
    a <- map read . words <$> getLine
    print $ solve a

maxim = 10^9 + 1
solve list = bSearch 0 maxim (judge list) 

judge ys x = invNum <= div (div (l*(l+1)) 2) 2
    where (invNum,_) = inversionNumber cs
          cs = convert x ys
          l = length ys
convert x = scanl (\acc y-> if y>=x then acc+1 else acc-1) 0

bSearch ok ng condition = if ng == ok + 1 
                          then ok
                          else if condition mid 
                            then bSearch mid ng condition
                            else bSearch ok mid condition
                        where mid = div (ok+ng) 2

merge_ [] right _ = (0,right)
merge_ left [] _ = (0,left)
merge_ left@(x:xs) right@(y:ys) leftLength = if x <= y  
                                  then (ansx,x:xList)
                                  else (leftLength + ansy, y:yList)
    where (ansy, yList) = merge_ left ys leftLength
          (ansx,xList) = merge_ xs right (leftLength-1)

inversionNumber [] = (0,[])
inversionNumber [x] = (0,[x])
inversionNumber xs = (a+b+c,merged)
    where left = take half xs
          right = drop half xs
          half = length xs `div` 2
          (a,ileft) = inversionNumber left
          (b,iright) = inversionNumber right
          (c,merged) = merge_ ileft iright half