思路：
- 题意：从[1，n]随机生成直角四面体的三边，求顶点到底面高h的 1/(n^2).
- 官方题解思路1：
- 官方题解思路2：
代码实现：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 998244353;
const int N = 6e6 + 5;
int t, n, a[N];
int qmi(int a, int k){
int res = 1;
while(k){
if(k & 1) res = (ll)res * a % mod;
k >>= 1;
a = (ll)a * a % mod;
}
return res;
}
void Inint(){
a[1] = 1;
for(int i = 2; i < N; i ++)
a[i] = (a[i-1] + qmi((i*i)%mod, mod-2))%mod;
}
signed main(){
IOS;
Inint();
cin >> t;
while(t --){
cin >> n;
cout << (a[n]*qmi(n, mod-2)%mod)*3%mod << endl;
}
return 0;
}