CF1794D.Counting Factorizations

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题目描述

The prime factorization of a positive integer mm is the unique way to write it as m=p1e1p2e2pkek\displaystyle m=p_1^{e_1}\cdot p_2^{e_2}\cdot \ldots \cdot p_k^{e_k} , where p1,p2,,pkp_1, p_2, \ldots, p_k are prime numbers, p1<p2<<pkp_1 < p_2 < \ldots < p_k and e1,e2,,eke_1, e_2, \ldots, e_k are positive integers.

For each positive integer mm , f(m)f(m) is defined as the multiset of all numbers in its prime factorization, that is f(m)={p1,e1,p2,e2,,pk,ek}f(m)=\{p_1,e_1,p_2,e_2,\ldots,p_k,e_k\} .

For example, f(24)={2,3,3,1}f(24)=\{2,3,3,1\} , f(5)={1,5}f(5)=\{1,5\} and f(1)={}f(1)=\{\} .

You are given a list consisting of 2n2n integers a1,a2,,a2na_1, a_2, \ldots, a_{2n} . Count how many positive integers mm satisfy that f(m)={a1,a2,,a2n}f(m)=\{a_1, a_2, \ldots, a_{2n}\} . Since this value may be large, print it modulo 998244353998\,244\,353 .

输入格式

The first line contains one integer nn ( 1n20221\le n \le 2022 ).

The second line contains 2n2n integers a1,a2,,a2na_1, a_2, \ldots, a_{2n} ( 1ai1061\le a_i\le 10^6 ) — the given list.

输出格式

Print one integer, the number of positive integers mm satisfying f(m)={a1,a2,,a2n}f(m)=\{a_1, a_2, \ldots, a_{2n}\} modulo 998244353998\,244\,353 .

输入输出样例

  • 输入#1

    2
    1 3 2 3

    输出#1

    2
  • 输入#2

    2
    2 2 3 5

    输出#2

    5
  • 输入#3

    1
    1 4

    输出#3

    0

说明/提示

In the first sample, the two values of mm such that f(m)={1,2,3,3}f(m)=\{1,2,3,3\} are m=24m=24 and m=54m=54 . Their prime factorizations are 24=233124=2^3\cdot 3^1 and 54=213354=2^1\cdot 3^3 .

In the second sample, the five values of mm such that f(m)={2,2,3,5}f(m)=\{2,2,3,5\} are 200,225,288,500200, 225, 288, 500 and 972972 .

In the third sample, there is no value of mm such that f(m)={1,4}f(m)=\{1,4\} . Neither 141^4 nor 414^1 are prime factorizations because 11 and 44 are not primes.

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