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

    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            #i & -iでLSB(least significant bit/最下位 bit)が求まる
            #区間和はLSBだけ引いていく
            i -= i & -i
        return s

    #a_i(1<=i<=n)にxを加算
    #点に対する更新
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

def invnumber(n, S):
    bit = Bit(2*n+1)
    invs = 0
    for i in range(n):
        s = S[i]+n #BITで扱うため
        invs += bit.sum(s) # #i<j
        bit.add(s, 1)
    return invs

n = int(input())
a = list(map(int, input().split()))
r = max(a)+1
l = 0
c = (n*(n+1)//2+1)//2
while r-l > 1:
    m = (l+r)//2
    S = [0]*(n+1)
    for i in range(1, n+1):
        if a[i-1] >= m:
            S[i] = S[i-1] + 1
        else:
            S[i] = S[i-1] - 1
    if invnumber(n+1, S) >= c:
        l = m
    else:
        r = m
print(l)
