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.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8
from itertools import accumulate
from math import ceil
N = int(input())
A = list(map(int, input().split()))


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 + 1)  # -N ~ +Nまで。0はN番目。
# にぶたん：「答えがX以上か？」
ok, ng = 0, 10 ** 9 + 1
while abs(ok - ng) > 1:
    X = (ok + ng) // 2

    B = [(1 if a >= X else - 1) for a in A]
    B = [0] + list(accumulate(B))

    cnt = 0
    BIT.reset()
    B_sorted = {b: i for i, b in enumerate(sorted(B), start=1)}
    for j, b in enumerate(B):
        cnt += BIT.sum(B_sorted[b])
        BIT.add(B_sorted[b], 1)

    if cnt >= ceil((N * (N + 1)) // 2 / 2):
        ok = X
    else:
        ng = X

print(ok)
