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).

Input

Input is given from Standard Input in the following format:
N
$X_1 X_2 \ldots X_N$

Output

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

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 $i\neq j$.

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 $\displaystyle \lim_{K\to\infty} \frac{f(K)}{K^3} = \frac{1}{24}$.

Sample Input 2

3
5 9 2

Sample Output 2

832860616

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

Sample Input 3

2
2 3

Sample Output 3

166374059

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

Sample Input 4

4
4 5 3 2

Sample Output 4

0

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

见题面 Input 部分

见题面 Output 部分