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
NN = 18
N = int(input())
A = [int(a) for a in input().split()]

def chk(k):
    s = 0
    c = 0
    BIT=[0]*(2**NN+1)
    def add(i, x=1):
        i += 1 << NN - 1
        while i <= 2**NN:
            BIT[i] += x
            i += i & (-i)

    def getsum(i):
        ret = 0
        i += 1 << NN - 1
        while i != 0:
            ret += BIT[i]
            i -= i&(-i)
        return ret
    add(0)
    for i in range(N):
        if A[i] >= k:
            s += 1
        else:
            s -= 1
        c += getsum(s)
        add(s)
    if 4 * c >= N * (N+1):
        return 1
    return 0

l, r = 0, 1<<30
while r - l > 1:
    m = (l+r) // 2
    if chk(m):
        l = m
    else:
        r = m
print(l)
