Codeforces Round 768 (Div. 1) |
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Finished |

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combinatorics

math

number theory

probabilities

*2900

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E. Expected Components

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputGiven a cyclic array $$$a$$$ of size $$$n$$$, where $$$a_i$$$ is the value of $$$a$$$ in the $$$i$$$-th position, there may be repeated values. Let us define that a permutation of $$$a$$$ is equal to another permutation of $$$a$$$ if and only if their values are the same for each position $$$i$$$ or we can transform them to each other by performing some cyclic rotation. Let us define for a cyclic array $$$b$$$ its number of components as the number of connected components in a graph, where the vertices are the positions of $$$b$$$ and we add an edge between each pair of adjacent positions of $$$b$$$ with equal values (note that in a cyclic array the first and last position are also adjacents).

Find the expected value of components of a permutation of $$$a$$$ if we select it equiprobably over the set of all the different permutations of $$$a$$$.

Input

The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$) — the size of the cyclic array $$$a$$$.

The second line of each test case contains $$$n$$$ integers, the $$$i$$$-th of them is the value $$$a_i$$$ ($$$1 \le a_i \le n$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

It is guaranteed that the total number of different permutations of $$$a$$$ is not divisible by $$$M$$$

Output

For each test case print a single integer — the expected value of components of a permutation of $$$a$$$ if we select it equiprobably over the set of all the different permutations of $$$a$$$ modulo $$$998\,244\,353$$$.

Formally, let $$$M = 998\,244\,353$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{M}$$$. Output the integer equal to $$$p \cdot q^{-1} \bmod M$$$. In other words, output such an integer $$$x$$$ that $$$0 \le x < M$$$ and $$$x \cdot q \equiv p \pmod{M}$$$.

Example

Input

541 1 1 141 1 2 141 2 1 254 3 2 5 1121 3 2 3 2 1 3 3 1 3 3 2

Output

1 2 3 5 358642921

Note

In the first test case there is only $$$1$$$ different permutation of $$$a$$$:

- $$$[1, 1, 1, 1]$$$ has $$$1$$$ component.
- Therefore the expected value of components is $$$\frac{1}{1} = 1$$$

In the second test case there are $$$4$$$ ways to permute the cyclic array $$$a$$$, but there is only $$$1$$$ different permutation of $$$a$$$:

- $$$[1, 1, 1, 2]$$$, $$$[1, 1, 2, 1]$$$, $$$[1, 2, 1, 1]$$$ and $$$[2, 1, 1, 1]$$$ are the same permutation and have $$$2$$$ components.
- Therefore the expected value of components is $$$\frac{2}{1} = 2$$$

In the third test case there are $$$6$$$ ways to permute the cyclic array $$$a$$$, but there are only $$$2$$$ different permutations of $$$a$$$:

- $$$[1, 1, 2, 2]$$$, $$$[2, 1, 1, 2]$$$, $$$[2, 2, 1, 1]$$$ and $$$[1, 2, 2, 1]$$$ are the same permutation and have $$$2$$$ components.
- $$$[1, 2, 1, 2]$$$ and $$$[2, 1, 2, 1]$$$ are the same permutation and have $$$4$$$ components.
- Therefore the expected value of components is $$$\frac{2+4}{2} = \frac{6}{2} = 3$$$

In the fourth test case there are $$$120$$$ ways to permute the cyclic array $$$a$$$, but there are only $$$24$$$ different permutations of $$$a$$$:

- Any permutation of $$$a$$$ has $$$5$$$ components.
- Therefore the expected value of components is $$$\frac{24\cdot 5}{24} = \frac{120}{24} = 5$$$

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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