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
from itertools import accumulate

class BIT_RSQ():
    def __init__(self, n):
        self.n = n
        self.data = [0]*(n+1)

    def add(self, i, v):
        while i <= self.n:
            self.data[i-1] += v
            i += i & -i

    def sum(self, i):
        ret = 0
        while(i > 0):
            ret += self.data[i-1]
            i -= i & -i
        return ret

n = int(input())
a = list(map(int, input().split()))

def check(m):
    b = [[-1,1][i>=m] for i in a]
    bit = BIT_RSQ(2*n+1)
    s = [0] + list(accumulate(b))
    res = 0
    for i in range(n):
        bit.add(s[i]+n+1, 1)
        res += bit.sum(s[i+1]+n+1)
    return res >= (n*(n+1)//2+1)//2

left = 0
right = 10**9+1
while right-left > 1:
    mid = (right+left)//2
    if check(mid):
        left = mid
    else:
        right = mid
print(left)