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
from itertools import accumulate
from math import ceil

N = int(input())
a = list(map(int, input().split()))

class BIT:
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (n + 1)
        
    def update(self, i, v):
        while i <= self.n:
            self.bit[i] += v
            i += i & -i
            
    def query(self, i):
        ret = 0
        while i > 0:
            ret += self.bit[i]
            i -= i & -i
        return ret
    
    def reset(self):
        self.bit = [0] * (self.n + 1)

ub = max(a) + 1
lb = 0
th = ceil(((N + 1) * N // 2) / 2)
bit = BIT(N + 1)

while ub - lb > 1:
    mid = (ub + lb) // 2
    b = [1 if v >= mid else -1 for v in a]
    b = [0] + list(accumulate(b))
    b_sorted = {v: i + 1 for i, v in enumerate(sorted(b))}
    
    # Count inversion
    cnt = 0
    bit.reset()
    for v in b:
        cnt += bit.query(b_sorted[v])
        bit.update(b_sorted[v], 1)
        
    if cnt >= th:
        lb = mid
    else:
        ub = mid
    
print(lb)





