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
input = sys.stdin.readline
from collections import *

class BIT:
    #n:要素数
    def __init__(self, n):
        self.n = n
        self.bit = [0]*(n+1)
    
    #i番目(0-indexed)の値にxを足す
    def add(self, i, x):
        i += 1
        
        while i<=self.n:
            self.bit[i] += x
            i += i&(-i)
    
    #0からi番目までの値の和を求める
    def acc(self, i):
        i += 1
        s = 0
        
        while i>0:
            s += self.bit[i]
            i -= i&(-i)
        
        return s
        
def compress(l):
    l = list(set(l))
    l.sort()
    idx = defaultdict(int)
    
    for i in range(len(l)):
        idx[l[i]] = i
    
    return idx

def judge(x):
    l = []
    
    for ai in a:
        if ai>=x:
            l.append(1)
        else:
            l.append(-1)
    
    acc = [0]*(N+1)
    
    for i in range(N):
        acc[i+1] = acc[i]+l[i]
    
    idx = compress(acc)
    bit = BIT(len(idx.keys()))
    cnt = 0
    
    for i in range(N+1):
        cnt += bit.acc(idx[acc[i]])
        bit.add(idx[acc[i]], 1)

    return cnt>=(N*(N+1)//2+1)//2
    
def binary_search():
    l, r = 0, max(a)
    
    while l<=r:
        m = (l+r)//2
        
        if judge(m):
            l = m+1
        else:
            r = m-1
    
    return r

N = int(input())
a = list(map(int, input().split()))
print(binary_search())