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 = [int(i) for i in input().split()]
AS = sorted(list(set(A)))

def f(x):
    B = [((x<=a)<<1)-1 for a in A]
    C = [N]
    [C.append(C[-1]+b) for b in B]

    r = 0
    Bit = [0] * (2*N+1)

    for i,c in enumerate(C):
        t=c-1
        while(t>0):
            r += Bit[t]
            t -= t&-t
        t=c-1
        while(t<=2*N+1):
            Bit[t] += 1
            t += t&-t
    return(r>=N*(N+1)//4)

l = 0
r = len(AS)-1
while l != r:
    if f(AS[(l+r+1)//2]):
        l = (l+r+1)//2
    else:
        r = (((l+r+1)//2)-1)
print(AS[r])
