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()))

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

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

    def add(self, i, x):
        while i <= self.size:
            self.bit[i] += x
            i += i & -i
ok = 0
ng = 10**9+1

def inversion(a, bit):
    ans = 0
    for i, p in enumerate(b):
        bit.add(p, 1)
        ans += i+1 - bit.sum(p)
    return ans

while ng - ok > 1:
    m = ok + (ng-ok)//2
    b = [0]
    for i in range(n):
        if a[i] >= m:
            b.append(b[i]+1)
        else:
            b.append(b[i]-1)
    b = [i+100001 for i in b]
    bit = BIT(200005)
    inversion = 0
    for i, p in enumerate(b):
        bit.add(p, 1)
        inversion += i+1 - bit.sum(p)

    if inversion <= n*(n+1)//4:
        ok = m
    else:
        ng = m



print(ok)
