#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc=998244353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(modc-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{
        return x == a.x;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){
        return x;
    }
};


int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    int N;
    cin >> N;
    //dp(i,j) = 左からi個、右からj個取り除く方法の数
    vector dp(N*2+1, vector<mint>(N*2+1));
    dp[0][0] = 1;

    for (int i=0; i<=N*2; i++){
        for (int j=0; j<=N*2; j++){
            if (2 <= i) dp[i][j] += dp[i-2][j] * ((i-1) * (i-2) / 2); //LL
            if (2 <= j) dp[i][j] += dp[i][j-2] * ((j-1) * (j-2) / 2); //RR
            if (1 <= i && 1 <= j) dp[i][j] += dp[i-1][j-1] * ((i-1) * (j-1) + 1); //LR
        }
    }

    for (int i=0; i<=N*2; i++) cout << dp[i][N*2-i] << endl;

    return 0;
}