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()))
m = (n * (n + 1) // 2) // 2 + 1


class BIT1():
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (self.n + 1)
        self.data = [0] * (self.n + 1)

    def add(self, idx, x):
        # add x to idx-th element
        # idx: 1-indexed
        self.data[idx] += x
        while idx <= self.n:
            self.bit[idx] += x
            idx += (idx & (-idx))

    def sum(self, idx):
        # get sum of [1, idx]
        # idx: 1-indexed
        s = 0
        while idx:
            s += self.bit[idx]
            idx -= (idx & (-idx))
        return s


def check(x):
    # MofM が x以下である
    # x以下の要素が (n*(n+1)//2)//2 + 1 個以上ある

    s = 0
    c = [0] * (n + 1)
    for i in range(n):
        c[i + 1] = c[i]
        if a[i] <= x:
            c[i + 1] += 1
    d = [2 * c[i] - i + n + 1 for i in range(n + 1)]
    d_max = max(d)
    bit = BIT1(d_max)

    cnt = 0
    for i in range(n + 1):
        cnt += bit.sum(d[i] - 1)
        bit.add(d[i], 1)
    return cnt >= m


lb = 0  # False
ub = max(a) + 1  # True
while ub - lb > 1:
    mid = (ub + lb) // 2
    if check(mid):
        ub = mid
    else:
        lb = mid

print(ub)
