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 = [int(x) for x in input().split()]

b = sorted(a)
M = ((N + 1) * N // 2 + 1) // 2


class BIT(object):
    def __init__(self, N=1):
        self.N = N
        self.bit = [0] * (N + 1)

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

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


def is_OK(x):
    S = [1] * N
    for i in range(N):
        if a[i] < x:
            S[i] = -1
        if i > 0:
            S[i] += S[i - 1]
    S = [0] + S
    min_S = min(S)
    max_S = max(S)

    bit = BIT(max_S - min_S + 1)
    cnt = 0
    for S_i in S:
        S_i = S_i - min_S + 1
        cnt += bit.sum(S_i)
        bit.add(S_i, 1)
    return cnt >= M


left, right = 0, 10 ** 9 + 1
while right - left > 1:
    mid = (left + right) // 2
    if is_OK(mid):
        left = mid
    else:
        right = mid

print(left)