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
from math import ceil
from itertools import accumulate
n = int(input())
import math
import decimal
import collections
import itertools
import sys
#Union-Find
def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]
def unite(x, y):
    p = find(x)
    q = find(y)
    if p == q:
        return None
    if p > q:
        p,q = q,p
    par[p] += par[q]
    par[q] = p
def same(x, y):
    return find(x) == find(y)
def size(x):
    return -par[find(x)]
par = [-1 for i in range(n)]

#素数関連
def prime_numbers(x):
    if x < 2:
        return []
    prime_numbers = [i for i in range(x)]
    prime_numbers[1] = 0
    for prime_number in prime_numbers:
        if prime_number > math.sqrt(x):
            break
        if prime_number == 0:
            continue
        for composite_number in range(2 * prime_number, x, prime_number):
            prime_numbers[composite_number] = 0
    return [prime_number for prime_number in prime_numbers if prime_number != 0]
def is_prime(x):
    if x < 2:
        return False
    if x == 2 or x == 3 or x == 5:
        return True
    if x % 2 == 0 or x % 3 == 0 or x % 5 == 0:
        return False
    prime_number = 7
    difference = 4
    while prime_number <= math.sqrt(x):
        if x % prime_number == 0:
            return False
        prime_number += difference
        difference = 6 - difference
    return True
#nCr
mod = 10 ** 9 + 7
def comb(n, r):
    if n < r:return 0
    if n < 0 or r < 0:return 0
    return fa[n] * fi[r] % mod * fi[n - r] % mod
fa = [1] * (n + 1)
fi = [1] * (n + 1)
for i in range(1, n + 1):
    fa[i] = fa[i - 1] * i % mod
    fi[i] = pow(fa[i], mod - 2, mod)
def add(i, x):
    while i <= len(BIT):
        BIT[i] += x
        i += i & -i
def query(i):
    s = 0
    while i > 0:
        s += BIT[i]
        i -= i & -i
    return s
a = list(map(int, input().split()))
lis = sorted(list(set(a)))
l, r = 0, len(lis)
while r - l > 1:
    BIT = [0] * (2 * (n + 1) + 100)
    m = (l + r) // 2
    cum = [0]
    for i in range(n):
        cur = 1 if a[i] >= lis[m] else -1
        cum.append(cum[-1] + cur)
    cnt = 0
    for i in range(n + 1):
        cur = n + cum[i] + 5
        cnt += query(cur)
        add(cur, 1)
    if n * (n + 1) // 2 <= cnt * 2:
        l = m 
    else:
        r = m
print(lis[l])