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 & -i
        return s

    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i


n = int(input())
A = list(map(int, input().split()))
all_pattern = n * (n+1) // 2
check = all_pattern // 2 + bool(all_pattern % 2)
X = sorted(A)

left = 0
right = n

while left + 1 < right:
    idx = (left+right) // 2
    x = X[idx]
    up_or_down = [1 if num >= x else -1 for num in A]
    cumsum = [n]
    total = n
    for point in up_or_down:
        total += point
        cumsum.append(total)

    tree = Bit(2*n+1)
    pattern = 0
    for p in cumsum:
        pattern += tree.sum(p)
        tree.add(p, 1)

    if pattern >= check:
        left = idx
    else:
        right = idx

print(X[left])
