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()))

if n == 1:
    print(a[0])
    exit()

b = list(sorted(set(a)))

def count_inversion(ans):
    s = [0 for _ in range(2 * n + 2) ]

    def bit_add(i):
        x = i
        while x < 2 * n:
            s[x] += 1
            x += x & -x

    def bit_sum(i):
        x = i
        ret = 0
        while x > 0:
            ret += s[x]
            x -= x & -x
        return ret

    temp = [n for _ in range(n + 1)]
    temp[0] = n + 1
    for i in range(n):
        if a[i] >= b[ans]:
            temp[i + 1] = temp[i] + 1
        else:
            temp[i + 1] = temp[i] - 1
    res = 0
    for i in range(n + 1):
        x = temp[i]
        res += bit_sum(x)
        bit_add(x)
    return res

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

ans = 0

l = 0
r = len(b)
while 1:
    ans = (l + r) // 2
    if count_inversion(ans) * 2 >= m:
        l = ans
    else:
        r = ans
    if r - l <= 1:
        ans = l
        break

print(b[ans])
