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
import math
 
n = int(input())
a = [int(i) for i in input().split(" ")]

m = n * (n + 1) // 2
 
class BIT:
    def __init__(self, n):
        self.t = [0] * (n + 1)
    
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.t[i]
            i -= (i & -i)
        return s
    
    def add(self, i, v):
        while i < len(self.t):
            self.t[i] += v
            i += (i & -i)
            
def count(s):
    s_min = min(s)
    b = BIT(max(s) - s_min + 1)
    t = [0] * len(s)
    for j in range(0, len(s)):
        t[j] = b.sum(s[j] - s_min + 1)
        b.add(s[j] - s_min + 1, 1)
    return sum(t)
            
def bs(l, r):
    if l == r:
        return l
    else:
        h = math.ceil((l + r) / 2)
        
        ab = [-1 if a[i] < h else 1 for i in range(0, n)]
        s = [0] * (n + 1)
        
        for i in range(1, n + 1):
            s[i] = s[i - 1] + ab[i - 1]
            
        if count(s) >= math.ceil(m / 2):
            return bs(h, r)
        else:
            return bs(l, h - 1)
        
print(bs(min(a), max(a)))