Codeforces Round #779 B. Marin and Anti-coprime Permutation

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 n is a permutation that has the following property:

$\gcd (1\cdot p_1,2\cdot p_2,\dots ,n\cdot p_n)>1,$

where$ \gcd $is the greatest common divisor.

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

Input

The first line contains one integer t$(1\le t\le 10^3)$ — the number of test cases.

Each test case consists of one line containing one integer $n(1\le n\le 10^3).$

Output

For each test case, print one integer — number of beautiful permutations. Because the answer can be very big, please print the answer modulo$ 998,244,353.$

Example

input

Copy

7
1
2
3
4
5
6
1000

output

Copy

0
1
0
4
0
36
665702330

Note

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

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

题目分析：

这道题比较巧妙，虽然说只要让gcd不唯一，其实醉酒当的方法就是让gcd为2，也就是全部变成偶数，所以只需要让排列中的奇数乘偶数，偶数乘奇数，这样分配，就能得到数量，如果排列有奇数个数，那么无法让拍列的奇数都成偶数，所以不可能成立，只有排列为偶数个，然后奇数位和偶数位分别求阶乘乘起来就是结果(求阶乘的话，在程序开始前求出来就好，记得取模)

#include<iostream>
using namespace std;
#define mod 998244353
typedef unsigned long long ll;
ll a[504];
int main()
{
    int n;
    cin>>n;
    int tmp;
    a[1]=1;
    for(int i=2;i<=500;i++){
        a[i]=a[i-1]*i%mod;
    }
    while(n--){
        cin>>tmp;
        if(tmp%2!=0)cout<<"0"<<endl;
        else cout<<a[tmp/2]*a[tmp/2]%mod<<endl;
    }
}