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 collections import defaultdict
class Bit:
  def __init__(self,n):
    self.size = n
    self.tree = [0]*(n+1)
 
  def sum(self,i):
    s = 0
    while i > 0:
      s += self.tree[i]
      i -= i & -i
    return s

  def add(self,i,x):
    while i <= self.size:
      self.tree[i] += x
      i += i & -i
n = int(input())
a = list(map(int,input().split()))
def jdg(x):
  b = [0 for i in range(n)]
  for i in range(n):
    if a[i] >= x:
      b[i] = 1
    else:
      b[i] = -1
  acc = [0]+list(accumulate(b))
  mn = min(acc)
  mx = max(acc)
  dif = mx-mn
  bit = Bit(dif+2)
  ret = 0
  for i in range(n+1):
    p = acc[i]-mn+1
    ret += bit.sum(p)
    bit.add(p,1)
  return ret
l = 0
r = max(a)+1
while l+1 < r:
  x = (l+r)//2
  if jdg(x) >= (n+1)*n//4:
    l = x
  else:
    r = x
print(l)