Median of an Array
You are given an array $$$a$$$ of $$$n$$$ integers.The median of an array $$$q_1, q_2, \ldots, q_k$$$ is the number $$$p_{\lceil \frac{k}{2} \rceil}$$$, where $$$p$$$ is the array $$$q$$$ sorted in non-decreasing order. For example, the median of the array $$$[9, 5, 1, 2, 6]$$$ is $$$5$$$, as in the sorted array $$$[1, 2, 5, 6, 9]$$$, the number at index $$$\lceil \frac{5}{2} \rceil = 3$$$ is $$$5$$$, and the median of the array $$$[9, 2, 8, 3]$$$ is $$$3$$$, as in the sorted array $$$[2, 3, 8, 9]$$$, the number at index $$$\lceil \frac{4}{2} \rceil = 2$$$ is $$$3$$$.You are allowed to choose an integer $$$i$$$ ($$$1 \le i \le n$$$) and increase $$$a_i$$$ by $$$1$$$ in one operation.Your task is to find the minimum number of operations required to increase the median of the array.Note that the array $$$a$$$ may not necessarily contain distinct numbers.
In the first test case, you can apply one operation to the first number and obtain the array $$$[3, 2, 8]$$$, the median of this array is $$$3$$$, as it is the number at index $$$\lceil \frac{3}{2} \rceil = 2$$$ in the non-decreasing sorted array $$$[2, 3, 8]$$$. The median of the original array $$$[2, 2, 8]$$$ is $$$2$$$, as it is the number at index $$$\lceil \frac{3}{2} \rceil = 2$$$ in the non-decreasing sorted array $$$[2, 2, 8]$$$. Thus, the median increased ($$$3 > 2$$$) in just one operation.In the fourth test case, you can apply one operation to each of the numbers at indices $$$1, 2, 3$$$ and obtain the array $$$[6, 6, 6, 4, 5]$$$, the median of this array is $$$6$$$, as it is the number at index $$$\lceil \frac{5}{2} \rceil = 3$$$ in the non-decreasing sorted array $$$[4, 5, 6, 6, 6]$$$. The median of the original array $$$[5, 5, 5, 4, 5]$$$ is $$$5$$$, as it is the number at index $$$\lceil \frac{5}{2} \rceil = 2$$$ in the non-decreasing sorted array $$$[4, 5, 5, 5, 5]$$$. Thus, the median increased ($$$6 > 5$$$) in three operations. It can be shown that this is the minimum possible number of operations.In the fifth test case, you can apply one operation to each of the numbers at indices $$$1, 3$$$ and obtain the array $$$[3, 1, 3, 3, 1, 4]$$$, the median of this array is $$$3$$$, as it is the number at index $$$\lceil \frac{6}{2} \rceil = 3$$$ in the non-decreasing sorted array $$$[1, 1, 3, 3, 3, 4]$$$. The median of the original array $$$[2, 1, 2, 3, 1, 4]$$$ is $$$2$$$, as it is the number at index $$$\lceil \frac{6}{2} \rceil = 3$$$ in the non-decreasing sorted array $$$[1, 1, 2, 2, 3, 4]$$$. Thus, the median increased ($$$3 > 2$$$) in two operations. It can be shown that this is the minimum possible number of operations.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. Then follows the description of the test cases.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of the array $$$a$$$.The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$.It is guaranteed that the sum of the values of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, output a single integer — the minimum number of operations required to increase the median of the array.
for i in range(int(input())):
    n = int(input())
    lis = list(map(int, input().split()))
    lis.sort()
    if n % 2 != 0:
        half = n // 2 + 1
        mid = lis[half - 1]
        new = lis[half:len(lis)]
        res = new.count(mid)
        print(res + 1)
    else:
        half = n // 2
        mid = lis[half - 1]
        new = lis[half:len(lis)]
        res = new.count(mid)
        print(res + 1)
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, a) = (readInt(), readList())
    a.sort()
    idx = (n - 1) // 2
    cnt = a[idx:].count(a[idx])
    return cnt
for _ in range(int(input())):
    print(solve())
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, a) = (readInt(), sorted(readList()))
    return a[(n - 1) // 2:].count(a[(n - 1) // 2])
for _ in range(int(input())):
    print(solve())