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: Time Limit Exceeded)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

N = int(input())
a = list(map(int, input().split()))
assert len(a) == N

# 解説を参照
# https://img.atcoder.jp/arc101/editorial.pdf

def mergesort(S, f, t):
    if t - f <= 1:
        return 0, S[f:t]

    n1, S1 = mergesort(S, f, (f + t) // 2)
    n2, S2 = mergesort(S, (f + t) // 2, t)
    T = []
    n = 0
    i = 0
    j = 0
    while i < len(S1) and j < len(S2):
        if S1[i] <= S2[j]:
            T.append(S1[i])
            i += 1
            n += len(S2) - j
        else:
            T.append(S2[j])
            j += 1
    if i < len(S1):
        T.extend(S1[i:])
    if j < len(S2):
        T.extend(S2[j:])
    return n1 + n2 + n, T

def inversion(S):
    """リストSの転倒数"""
    n, __ = mergesort(S, 0, len(S))
    return n

def n_median_larger_than_or_equal_to(x):
    """aの部分列の中央値のうちx以上のものの個数"""

    # aの要素を -1, 1 に置換したリスト
    b = [-1 if e < x else 1 for e in a]
    # bの累積和
    S = [0]
    for e in b:
        S.append(S[-1] + e)
    # Sの転倒数
    return inversion(S)

# 条件を満たす x の最大値を2分探索で探す
lo = 1
hi = max(a)

ncomb = N * (N + 1) // 2    # 部分列の個数

while lo < hi:
    mid = (lo + hi + 1) // 2
    nm = n_median_larger_than_or_equal_to(mid)
    if nm >= (ncomb + 1) // 2:
        lo = mid
    else:
        hi = mid - 1

print(lo)