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(_) for _ in input().split()]


class BIT:
    # index 1~
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)

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

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


M = N * (N + 1) // 2  # (N+1)C2
sorted_A = sorted(A)

inf = 0
sup = max(A) + 1
while sup - inf > 1:
    x = (sup + inf) // 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

    min_value = min(S)
    for i in range(N + 1):
        S[i] += abs(min_value) + 1

    forward = 0
    bit = BIT(max(S))
    for s in S:
        forward += bit.sum(s)
        bit.add(s, 1)

    if forward >= (M + 1) // 2:
        inf = x
    else:
        sup = x

for a in reversed(sorted_A):
    if inf >= a:
        print(a)
        break
