B. Marin and Anti-coprime Permutation

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本文解析了Codeforces B题“Marin and Anti-coprime Permutation”，介绍了如何计算特定条件下美丽的排列数量，并提供了代码实现。重点在于理解题目要求的排列属性及通过数学方法求解。

Problem - B - Codeforces

B. Marin and Anti-coprime Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Marin wants you to count number of permutations that are beautiful. A beautiful permutation of length nn is a permutation that has the following property:

gcd(1⋅p1,2⋅p2,…,n⋅pn)>1,gcd(1⋅p1,2⋅p2,…,n⋅pn)>1,

where gcdgcd is the greatest common divisor.

A permutation is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation, but [1,2,2][1,2,2] is not a permutation (22 appears twice in the array) and [1,3,4][1,3,4] is also not a permutation (n=3n=3 but there is 44 in the array).

Input

The first line contains one integer tt (1≤t≤1031≤t≤103) — the number of test cases.

Each test case consists of one line containing one integer nn (1≤n≤1031≤n≤103).

Output

For each test case, print one integer — number of beautiful permutations. Because the answer can be very big, please print the answer modulo 998244353998244353.

Example

input

Copy

output

Copy

Note

In first test case, we only have one permutation which is [1][1] but it is not beautiful because gcd(1⋅1)=1gcd(1⋅1)=1.

In second test case, we only have one beautiful permutation which is [2,1][2,1] because gcd(1⋅2,2⋅1)=2gcd(1⋅2,2⋅1)=2.

from calendar import c
import sys
import math
#import numpy as np
#sys.stdin = open("in.txt", "r")
 
T = int(input())
M= 998244353
for _ in range(T):
    n= int(input())
    if n%2==1:
        print(0)
    else:
        a=math.factorial(n//2)%M
        b=(a*a)%M
        print(b)
    #break

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