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
def main():
    from itertools import accumulate as ac
    n = int(input())
    a = list(map(int, input().split()))

    class BIT():
        def __init__(self, n):
            self.n = n
            self.bit = [0]*(n+1)

        # 位置iに値vを足す
        def add(self, i, v):
            x = i+1
            while x < self.n + 1:
                self.bit[x] += v
                x += x & (-x)

        # 位置0からiまでの和(sum(bit[:i]))を計算する
        def sum(self, i):
            ret = 0
            x = i
            while x > 0:
                ret += self.bit[x]
                x -= x & (-x)
            return ret

        # 位置iからjまでの和(sum(bit[i:j]))を計算する
        def sum_range(self, i, j):
            return self.sum(j) - self.sum(i)

    def value(v):
        if n == 1:
            if v <= a[0]:
                return True
            else:
                return False
        b = [0]+list(ac([(i >= v)*2-1 for i in a]))
        bit = BIT(2*n+1)
        ans = 0
        bit.add(b[-1]+n, 1)
        for i in range(n-1, -1, -1):
            w = b[i]
            ans += bit.sum_range(w+n, 2*n+1)
            bit.add(w+n, 1)
        if ans >= (n*(n+1)//2)//2:
            return True
        else:
            return False

    def b_search(ok, ng, value):
        while abs(ok-ng) > 1:
            mid = (ok+ng)//2
            if value(mid):
                ok = mid
            else:
                ng = mid
        return ok

    print(b_search(0, 10**9+1, value))


main()
