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()]


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

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

    forward = 0
    size = max(S)
    bit = [0] * (size + 1)
    for s in S:
        i = s
        while i > 0:
            forward += bit[i]
            i -= i & -i

        i = s
        while i <= size:
            bit[i] += 1
            i += i & -i

    return forward >= (M + 1) // 2


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

inf = 0
sup = N
while sup - inf > 1:
    mid = (sup + inf) // 2
    x = sorted_A[mid]

    if sub_problem(x):
        inf = mid
    else:
        sup = mid

print(sorted_A[inf])
