Problem Statement

Given is a sequence of $N$ positive integers $X = (X_1, X_2, \\ldots, X_N)$.

For a positive integer $K$, let $f(K)$ be the number of triples of integers $(a,b,c)$ that satisfy all of the following conditions.

• $1\\leq c \\leq K$.
• $0\\leq a < c$ and $0\\leq b < c$.
• For each $i$, let $Y_i$ be the remainder when $aX_i + b$ is divided by $c$. Then, $Y_1 < Y_2 < \\cdots < Y_N$ holds.

It can be proved that the limit $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3}$ exists. Find this value modulo $998244353$ (see Notes).

Notes

We can prove that the limit in question is always a rational number. Additionally, under the Constraints of this problem, when that number is represented as $\\frac{P}{Q}$ using two coprime integers $P$ and $Q$, we can prove that there is a unique integer $R$ such that $R\\times Q\\equiv P\\pmod{998244353}$ and $0\\leq R < 998244353$. Find this $R$.

Constraints

• $2\\leq N\\leq 10^3$
• $X_i$ are positive integers such that $\\sum_{i=1}^N X_i \\leq 5\\times 10^5$.
• $X_i\\neq X_j$ if $i\\neq j$.

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Input

Input is given from Standard Input in the following format:

$N$ $X_1$ $X_2$ $\\ldots$ $X_N$

Output

Print $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3}$ modulo $998244353$.

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Sample Input 1

3
3 1 2

Sample Output 1

291154603

• For example, when $(a,b,c) = (3,5,7)$, we have $Y_1 = 0$, $Y_2 = 1$, $Y_3 = 4$, which satisfy $Y_1 < Y_2 < Y_3$.
• We have $f(1) = 0$, $f(2) = 0$, $f(3) = 1$, $f(4) = 2$, $f(5) = 5$.
• We have $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3} = \\frac{1}{24}$.

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Sample Input 2

3
5 9 2

Sample Output 2

832860616

We have $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3} = \\frac{55}{1008}$ .

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Sample Input 3

2
2 3

Sample Output 3

166374059

We have $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3} = \\frac{1}{6}$.

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Sample Input 4

4
4 5 3 2

Sample Output 4

0

We have $\\displaystyle \\lim_{K\\to\\infty} \\frac{f(K)}{K^3} = 0$.