Codeforces Round 856 (Div. 2)
D
Counting Factorizations
Codeforces
1900
原题
← CE →
翻译中…

The prime factorization of a positive integer $m$ is the unique way to write it as $\displaystyle m=p_1^{e_1}\cdot p_2^{e_2}\cdot \ldots \cdot p_k^{e_k}$, where $p_1, p_2, \ldots, p_k$ are prime numbers, $p_1 \lt p_2 \lt \ldots \lt p_k$ and $e_1, e_2, \ldots, e_k$ are positive integers.

翻译中…

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

翻译中…

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

翻译中…

You are given a list consisting of $2n$ integers $a_1, a_2, \ldots, a_{2n}$. Count how many positive integers $m$ satisfy that $f(m)=\{a_1, a_2, \ldots, a_{2n}\}$. Since this value may be large, print it modulo $998\,244\,353$.

翻译中…

Input

翻译中…

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

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The second line contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($1\le a_i\le 10^6$)  — the given list.

翻译中…

Output

翻译中…

Print one integer, the number of positive integers $m$ satisfying $f(m)=\{a_1, a_2, \ldots, a_{2n}\}$ modulo $998\,244\,353$.

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Note

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

翻译中…

In the second sample, the five values of $m$ such that $f(m)=\{2,2,3,5\}$ are $200, 225, 288, 500$ and $972$.

翻译中…

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

翻译中…

样例

Input #1
2
1 3 2 3
Output #1
2
InputCopy OutputCopy
Input #2
2
2 2 3 5
Output #2
5
InputCopy OutputCopy
Input #3
1
1 4
Output #3
0
InputCopy OutputCopy
题目信息
题号
1794D
时间限制
4 s
内存限制
256 MB
标签
组合数学
分治
动态规划
数学
数论

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