Before her final sortie, Chtholly asks Willem three questions.
The third is this: when the end finally comes, who will remain to witness it?
Willem cannot answer her directly. Instead, he draws a circle on the board, calling it the ring of all things, and writes down $$$n$$$ labeled integers $$$a_1,a_2,\\ldots,a_n$$$. Every possible order around the ring describes a different way in which the world might reach its end.
Consider a permutation $$$p_1,p_2,\\ldots,p_n$$$ of the integers from $$$1$$$ to $$$n$$$. Place the corresponding numbers on a circle in this order. The weight of the resulting circular arrangement is
$$$$$$ \\prod_{i=1}^{n}(a_{p_i}+a_{p_{i+1}}), $$$$$$
where $$$p_{n+1}=p_1$$$.
Two permutations describe the same circular arrangement if one can be obtained from the other by a cyclic shift. Reversing an arrangement does not make it the same arrangement; in other words, reflected arrangements are considered different unless they also coincide after a cyclic shift.
Find the sum of the weights of all distinct circular arrangements. Since the answer may be large, output it modulo $$$998\\,244\\,353$$$.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.
The first line of each test case contains one integer $$$n$$$ ($$$3 \\le n \\le 2\\cdot 10^5$$$) — the number of labeled integers.
The second line contains $$$n$$$ integers $$$a_1,a_2,\\ldots,a_n$$$ ($$$0 \\le a_i \\lt 998\\,244\\,353$$$).
It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\\cdot 10^5$$$.
For each test case, output one integer — the sum of the weights of all distinct circular arrangements, modulo $$$998\\,244\\,353$$$.
331 2 360 1 0 1 0 110114514 1919810 350234 11831 314159265 271828182 123456789 998244352 5201314 23333333
12012265885269
In the first test case, there are two distinct circular arrangements. They can be represented by the permutations $$$[1,2,3]$$$ and $$$[1,3,2]$$$. Both have weight
$$$$$$ (1+2)(2+3)(3+1)=60, $$$$$$
so the answer is $$$120$$$.
In the second test case, an arrangement has nonzero weight only if zeros and ones alternate around the circle. There are
$$$$$$ \\frac{2\\cdot3!\\cdot3!}{6}=12 $$$$$$
such circular arrangements: the factor $$$2$$$ chooses whether a linear representative starts with a zero or a one, and division by $$$6$$$ identifies cyclic shifts. Each arrangement has weight $$$1$$$. All other arrangements have weight $$$0$$$, so the answer is $$$12$$$.