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()))
x = list(sorted(set(a)))
n_sub = n * (n + 1) // 2
Li = 0
Ri = len(x)
while Ri > Li + 1:
    i = (Li + Ri) // 2
    xi = x[i]
    UorD = [1 if ak >= xi else -1 for ak in a]
    cnt_sum = [0] * (2 * n + 1)
    cumsum = 0
    cnt_sum[cumsum + n] = 1
    judge = 0
    s = 0
    for uk in UorD:
        cumsum += uk
        if uk == 1:
            s += cnt_sum[cumsum + n] + 1
        else:
            s -= cnt_sum[cumsum + n + 1] - 1
        cnt_sum[cumsum + n] += 1
        judge += s
        if judge >= n_sub // 2:
            Li = i
            break
    else:
        Ri = i
print(x[Li])
