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.N = n
        self.n = 1<<n.bit_length()
        self.data = [0] * (self.n + 1)
        
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.data[i]
            i -= i & -i
        return s

    def add(self, i, x):
        while i <= self.n:
            self.data[i] += x
            i += i & -i
N = int(input())
A = list(map(int, input().split()))
N0 = N*(N+1)//2
def C(x):
    B = BIT(N*2+1)
    s = [0]*(N+1)
    for i in range(N):
        a = A[i]
        s[i] = 1 if x<a else -1
        s[i]+=s[i-1]
    res = 0
    for i in range(-1, N):
        res+=B.sum(s[i]+N+1)
        B.add(s[i]+N+1, 1)
    return res<N0//2
ok = max(A)
ng = min(A)-1
while abs(ok-ng)>1:
    mid = (ok+ng)//2
    if C(mid):#midが条件を満たすか
        ok = mid
    else:
        ng = mid
print(ok)

    

    