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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from itertools import accumulate

N = int(input())
A = list(map(int, input().split()))

if N == 1:
    print(A[0])
    exit()


class BinaryIndexedTree:
    def __init__(self, n):
        self.size = n
        self.bit = [0] * (n + 1)

    def sum(self, i):
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= (i & -i)
        return s

    def add(self, i, x):
        while i <= self.size:
            self.bit[i] += x
            i += (i & -i)

    def reset(self):
        self.bit = [0] * (self.size + 1)


BIT = BinaryIndexedTree(N)  # Binaryの先頭の分
lo, hi = min(A) - 1, 10 ** 9 + 1
while hi - lo > 1:
    # Xは中央値候補
    X = (hi + lo) // 2

    # X以上なら1、X未満なら-1に変換
    Binary = [(1 if a >= X else -1) for a in A]

    # 累積和をとる
    Binary = list(accumulate(Binary))

    # BITを初期化
    BIT.reset()

    # 転倒数を数える
    inversion = 0
    Binary_sorted = {b: i for i, b in enumerate(sorted(Binary), start=1)}
    for j, b in enumerate(Binary):
        inversion += j - BIT.sum(Binary_sorted[b])
        BIT.add(Binary_sorted[b], 1)

    # 転倒数の割合で分岐
    if inversion / (N * (N + 1) / 2) < 0.5:
        lo = X
    else:
        hi = X

print(lo)