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
import sys
readline = sys.stdin.readline

class BIT:
    #1-indexed
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
        self.p = 2**(n.bit_length() - 1)
        self.dep = n.bit_length()
    def get(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
    
    def bl(self, v):
        if v <= 0:
            return -1
        s = 0
        k = self.p
        for _ in range(self.dep):
            if s + k <= self.size and self.tree[s+k] < v:
                s += k
                v -= self.tree[s+k]
            k //= 2
        return s + 1

def compress(L):
    L2 = list(set(L))
    L2.sort()
    C = {v : k for k, v in enumerate(L2)}
    return L2, C

N = int(readline())
A = list(map(int, readline().split()))
A2, C = compress(A)
A = [C[a] for a in A]
ok = 0
ng = max(A)+1
T = BIT(ng)
ma = min(A)+1
while abs(ok-ng) > 1:
    med = (ok+ng)//2
    B = [1 if a >= med else -1 for a in A]
    for i in range(1, N):
        B[i] += B[i-1]
    
    
    mb = -min(0, min(B))+1
    T = BIT(max(B)+mb)
    T.add(mb, 1)
    res = 0
    for b in B:
        res += T.get(mb+b)
        T.add(mb+b, 1)
    if res >= N*(N+1)//4:
        ok = med
    else:
        ng = med
print(A2[ok])