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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import numpy as np
N = int(input())
A = np.array(input().split(),dtype=np.int32)
total_medians = N*(N+1)//2
# n件のデータ
# (上側中央値 >= x) iff (a > x となる a が(n-1)//2以下)

def BIT_update(BIT,x,value):
  # 番号xにvalueを加える
  L = len(BIT)
  while x < L:
    BIT[x] += value
    bottom_bit = x & (-x)
    x += bottom_bit
  return

def BIT_sum(BIT,x):
  # 番号1～xの和を求める
  result = 0
  while x:
    result += BIT[x]
    bottom_bit = x & (-x)
    x -= bottom_bit
  return result

def test(x):
  # 「答はx以下である」
  # 「xより真に大きな区間中央値が(total_medians-1)//2個以下」
  B = np.concatenate([[0],(A > x).cumsum()])
  # count L,R such that -1<=L<R<=N-1 and B(R)-B(L) >= (R-L)/2
  # count L,R such that B(R) - R/2 >= B(L) - L/2
  C = 2*B-np.arange(N+1)
  C -= np.min(C) - 1
  M = C.max()
  BIT = [0]*(M+1)
  
  cnt = 0
  prev = C[0]
  for c in C[1:]:
    BIT_update(BIT,prev,1)
    cnt += BIT_sum(BIT,c) # 右端がRとなる(L,R)対
    prev = c
  return cnt <= (total_medians-1)//2

AA = sorted(set(A))
left = -1 # 答はAA[left]より大きい
right = len(AA)-1 # 答はAA[right]以下である
while right - left > 1:
  mid = (left+right)//2
  if test(AA[mid]):
    right = mid
  else:
    left = mid
    
answer = AA[right]
print(answer)