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()))
a2=sorted(set(a))
L=len(a2)
if n==1:print(a[0]);exit()
elif n==2:print(a[1]);exit()
ng=L
ok=0
while ng-ok>1:
    check=a2[(ok+ng)//2]
    cnt=0 #check以上の要素の数
    idx=n
    cl=[0]*(2*n+1)
    cl[n]=1
    Ru=0
    for i in range(n):
        if a[i]<check:
            Ru-=cl[idx]-1
            idx-=1
        else:
            idx+=1
            Ru+=cl[idx]+1
        cnt += Ru
        cl[idx]+=1
    if cnt >= ((n+1)*n//2)//2:
        ok = (ok+ng)//2
    else:
        ng = (ok+ng)//2

print(a2[ok])