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.tree = [0]*(n+1)

    def sum(self, i):
        # [0, i) の要素の総和を返す
        s = 0
        while i>0:
            s += self.tree[i]
            i -= i & -i
        return s

    def add(self, i, x):
        if not (0 <= i < self.size): raise ValueError("error!")
        i += 1
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

def search(x):
    l = [1 if i >= x else -1 for i in a]
    c = [n]
    for i in range(n):
        c.append(c[-1]+l[i])

    count = 0
    bit = Bit(2*n+1)
    for i in c:
        count += bit.sum(i+1)
        bit.add(i,1)
    
    return x,count

k = (n*(n+1)/2+1)//2
le = 0
ri = max(a)+1

while ri > le + 1:
    m = (ri+le)//2
    ans,count = search(m)

    if count >= k:
        le = m
    else:
        ri = m
print(le)
