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
# seishin.py
N = int(input())
*A, = map(int, input().split())

def solve(x):
    data = [0]*(N+1)
    def add(k, x):
        while k <= N:
            data[k] += x
            k += k & -k
    def get(k):
        s = 0
        while k:
            s += data[k]
            k -= k & -k
        return s
    c = 0; d = 10**18
    for a in A:
        if x <= a:
            c += 1
        else:
            c -= 1
        d = min(c, d)
    res = N*(N+1)//2
    c = -d+1
    for a in A:
        res -= get(c)
        if x <= a:
            c += 1
        else:
            c -= 1
        add(c, 1)
    return res <= N*(N+1)//4

left = min(A); right = max(A)+1
while left+1 < right:
    mid = (left + right) // 2
    if solve(mid):
        left = mid
    else:
        right = mid
print(left)
