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()))

def check(x):
    b = n
    r = 0
    y = 0
    D = [0]*(2*n+1)
    for i in range(n):
        D[b] += 1
        if a[i] < x:
            r += D[b]
            b += 1
        else:
            b -= 1
            r -= D[b]
        y += r
    return y

alpha = sorted(a)
l, r = 0, n
m, c = n//2, n*(n+1)//2
while True:
    if check(alpha[m]) <= c//2:
        if m == n-1:
            break
        elif check(alpha[m+1]) > c//2:
            break
        else:
            l, m = m, (m+r)//2
    else:
        m, r = (m+l)//2, m+1
        
print(alpha[m])