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
class BIT:
    def __init__(self,n):
        self.size=n
        self.tree=[0]*-~n
    def sum(self,i):
        s=0
        while i:
            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,*a=map(int,open(0).read().split())
m=n*-~n//2+1>>1
ng=max(a)+1
ok=0
while ng-ok>1:
    mid=(ok+ng)//2
    b=BIT(n-~n)
    c=n+1
    s=0
    b.add(c,1)
    for t in a:
        c+=1if t>=mid else-1
        s+=b.sum(c)
        b.add(c,1)
    if s>=m:ok=mid
    else:ng=mid
print(ok)