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 = [int(i) for i in input().split()]

half = ((N * (N + 1) // 2) + 1) // 2

class BinaryIndexedTree() :
    def __init__(self, N) :
        self.N = 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.N :
            self.tree[i] += x
            i += i & -i

def check(x) :
    b = [0] + [1 if a[i] >= x else -1 for i in range(N)]
    for i in range(1, N + 1) :
        b[i] += b[i-1]
    BIT = BinaryIndexedTree(2*N+2)
    
    ret = 0
    for i in range(N+1) :
        ret += BIT.sum(N+1+b[i])
        BIT.add(N+1+b[i], 1)
    
    return ret >= half
    
aa = sorted(a)
ok = 0
ng = N
while ng - ok > 1 :
    mid = (ok + ng) // 2
    if check(aa[mid]) :
        ok = mid
    else :
        ng = mid

print(aa[ok])
    