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 median_of_medians(N: int, A: list)->int:
    def search(c):
        bar = N
        r = 0
        res_nega = 0
        dp_arr = [0]*(2*N+1)
        for i in range(N):
            dp_arr[bar] += 1
            if A[i] < c:
                r += dp_arr[bar]
                bar += 1
            else:
                r -= (dp_arr[bar-1])
                bar -= 1
            res_nega += r
        return res_nega

    # binary search
    a_sorted = sorted(A)
    left = 0
    right = N
    mid = N // 2
    C = N*(N+1) // 2
    while True:
        if search(a_sorted[mid]) <= C // 2:
            if mid == N-1:
                break
            elif search(a_sorted[mid+1]) > C // 2:
                break
            else:
                left = mid
                mid = (mid + right) // 2
        else:
            right = mid + 1
            mid = (mid + left) // 2

    return a_sorted[mid]


if __name__ == "__main__":
    N = int(input())
    A = [int(s) for s in input().split()]
    ans = median_of_medians(N, A)
    print(ans)
