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
# 足す時は0~iまで一律に足し、返すのはi番目の値
class imosBIT():
    def __init__(self, N):
        self.N = N
        self.bit = [0 for _ in range(self.N+1)]
    
    def __str__(self):
        ret = []
        for i in range(1, self.N+1):
            ret.append(self.__getitem__(i))
        return "[" + ", ".join([str(a) for a in ret]) + "]"

    def __getitem__(self, i):
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s

    def add(self, i, x):
        while i <= self.N:
            self.bit[i] += x
            i += i & -i


import sys
input = sys.stdin.readline

N = int(input())
A = list(map(int, input().split()))

l = 0
r = max(A) + 1
while r-l > 1:
    x = (l+r)//2
    P = [0]
    for a in A:
        p = +1 if a >= x else -1
        P.append(P[-1]+p)
    score = 0
    bit = imosBIT(2*N+5)
    for p in P:
        p += N+1
        score += bit[p]
        bit.add(p, 1)
    
    if score >= (N*(N+1)//2+1)//2:
        l = x
    else:
        r = x

print(l)