Wish I Knew How to Sort
You are given a binary array $$$a$$$ (all elements of the array are $$$0$$$ or $$$1$$$) of length $$$n$$$. You wish to sort this array, but unfortunately, your algorithms teacher forgot to teach you sorting algorithms. You perform the following operations until $$$a$$$ is sorted:  Choose two random indices $$$i$$$ and $$$j$$$ such that $$$i < j$$$. Indices are chosen equally probable among all pairs of indices $$$(i, j)$$$ such that $$$1 \le i < j \le n$$$.  If $$$a_i > a_j$$$, then swap elements $$$a_i$$$ and $$$a_j$$$. What is the expected number of such operations you will perform before the array becomes sorted?It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{998\,244\,353}$$$. Output the integer equal to $$$p \cdot q^{-1} \bmod 998\,244\,353$$$. In other words, output such an integer $$$x$$$ that $$$0 \le x < 998\,244\,353$$$ and $$$x \cdot q \equiv p \pmod{998\,244\,353}$$$.
Consider the first test case. If the pair of indices $$$(2, 3)$$$ will be chosen, these elements will be swapped and array will become sorted. Otherwise, if one of pairs $$$(1, 2)$$$ or $$$(1, 3)$$$ will be selected, nothing will happen. So, the probability that the array will become sorted after one operation is $$$\frac{1}{3}$$$, the probability that the array will become sorted after two operations is $$$\frac{2}{3} \cdot \frac{1}{3}$$$, the probability that the array will become sorted after three operations is $$$\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3}$$$ and so on. The expected number of operations is $$$\sum \limits_{i=1}^{\infty} \left(\frac{2}{3} \right)^{i - 1} \cdot \frac{1}{3} \cdot i = 3$$$.In the second test case the array is already sorted so the expected number of operations is zero.In the third test case the expected number of operations equals to $$$\frac{75}{4}$$$ so the answer is $$$75 \cdot 4^{-1} \equiv 249\,561\,107 \pmod {998\,244\,353}$$$.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). Description of the test cases follows.The first line of each test case contains an integer $$$n$$$ ($$$1 \le n \le 200\,000$$$) — the number of elements in the binary array.The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i \in \{0, 1\}$$$) — elements of the array.It's guaranteed that sum of $$$n$$$ over all test cases does not exceed $$$200\,000$$$.
For each test case print one integer — the value $$$p \cdot q^{-1} \bmod 998\,244\,353$$$.
numOfTestcases = int(input())
ans = []

def helper(sumA, n, i):
    exp = pow(sumA - i, -2, 998244353)
    den = n * (n - 1)
    den //= 2
    return exp * den

def sol():
    n = int(input())
    a = [int(i) for i in input().split()]
    sumA = sum(a)
    partition = a[-sumA:]
    lastSumSum = sum(partition)
    count = 0
    for i in range(lastSumSum, sumA):
        count += helper(sumA, n, i)
        count %= 998244353
    ans.append(count)
for _ in range(numOfTestcases):
    sol()
for a in ans:
    print(a)
for _ in range(int(input())):
    n = int(input())
    a = [int(i) for i in input().split()]
    k = sum(a)
    st = sum(a[-k:])
    ans = sum((n * (n - 1) // 2 * pow(k - i, -2, 998244353) for i in range(st, k))) % 998244353
    print(ans)
t = int(input())
for test in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    s = sum(a)
    st = sum(a[-s:])
    ans = sum((n * (n - 1) // 2 * pow(s - i, -2, 998244353) for i in range(st, s))) % 998244353
    print(ans)