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 = list(map(int, input().split()))
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_list[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_list)
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
print(a_sorted[mid])
