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


class BIT:
    def __init__(self, n):
        self.n = n
        self.data = [0] * n

    def add(self, k, x):
        while k < self.n:
            self.data[k] += x
            k += k & -k

    def get(self, k):
        s = 0
        while k > 0:
            s += self.data[k]
            k -= k & -k

        return s

    def init(self, n):
        self.n = n
        self.data = [0] * n


left, right = 0, 10 ** 9 + 5
bit = BIT(N)
while right - left > 1:
    x = (right + left) // 2

    S = [0] * (N + 1)
    for i in range(N):
        if A[i] >= x:
            S[i + 1] = S[i] + 1
        else:
            S[i + 1] = S[i] - 1

    p = abs(min(S)) + 1
    for i in range(N + 1):
        S[i] += p
    q = max(S) + 1
    bit.init(q)
    num = 0
    for i in S:
        num += bit.get(i)
        bit.add(i, 1)

    if num * 2 >= N * (N + 1) // 2:
        left = x
    else:
        right = x

print(left)
