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():
    N, *A = map(int, open(0).read().split())
 
    B = sorted(A)
    l, r = 0, N
    m, c = N // 2, N * (N + 1) // 2
 
    def check(x):
        b, r, y = N, 0, 0
        D = [0] * (2 * N + 1)
        for a in A:
            D[b] += 1
            if a < x:
                r += D[b]
                b += 1
            else:
                b -= 1
                r -= D[b]
            y += r
        return y
 
    while True:
        if check(B[m]) <= c // 2:
            if m == N - 1 or check(B[m + 1]) > c // 2:
                break
            else:
                l, m = m, (m + r) // 2
        else:
            m, r = (m + l) // 2, m + 1
 
    print(B[m])
 
 
if __name__ == '__main__':
    main()