#pragma GCC optimize("Ofast") 
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
const int MOD = 998244353;
int mult(int a,int b)
{
	return (a*1LL*b)%MOD;
}
int qp(ll x, int y) {
    int z = 1;
    for (; y; y >>= 1, x = x * x % MOD) {
        if (y & 1) z = 1LL * z * x % MOD;
    }
    return z;
}
int inv(int a){return qp(a,MOD-2);}
int in[7];
const int MAXN = 3e5+5;
map<int,int>dp[MAXN];
int solve(int n,int k){
	if(n==1){
		//Bcout<<n<<" "<<k<<" "<<1<<'\n';
		return 1;
	}
	if(dp[n].find(k) != dp[n].end())return dp[n][k];
	int num = min(n,6);
	dp[n][k] = 0;
	for(int i=1;i<=num;i++){
		if(k%6 == i%6)continue;
		int k0 = k - (k-1)/6;
		int n0 = n - (n-1)/6;
		
		int rk = k%6;
		int rn = n%6;
		if(rk == 0)rk = 6;
		if(rn == 0)rn = 6;

		if(rk >= i)k0--;
		if(rn >= i)n0--;
		dp[n][k] += solve(n0,k0);
		if(dp[n][k]>=MOD)dp[n][k]-=MOD;
	}
	dp[n][k] = mult(dp[n][k],in[num]);
	//cout<<n<<" "<<k<<" "<<dp[n][k]<<'\n';
	return dp[n][k];
}
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=6;i++)in[i] = inv(i);
	for(int i=1;i<=n;i++)cout<<solve(n,i)<<'\n';


}