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(map(int,input().split()))
sortA=sorted(set(a))+[float('inf')]
def isOK(m):
  x=sortA[m]
  cnt=[0]*(2*n+1)
  cnt[n+1]=1
  i=n+1
  s=1
  res=0
  for b in a:
    if b>=x:
      i+=1
      s+=cnt[i]
    else:
      s-=cnt[i]
      i-=1
    res+=s
    cnt[i]+=1
    s+=1
  if res>=n*(n+1)//4:
      return False
  else:
      return True
l,r=0,len(sortA)-1
while r-l>1:
  mid=(r+l)//2
  if isOK(mid):
    r=mid
  else:
    l=mid
print(sortA[l])