Let \(q\) be the number of distinct integers that have appeared, and consider managing the following quantities for each value of \(q\):

• The probability that it is the first/second player’s turn right after the number of distinct integers has become \(q\), denoted as \(h_1, h_2\)
• The expected amount of the fine paid by the first/second player at the moment right after the number of distinct integers has become \(q\), denoted as \(f_1, f_2\)

The problem can be solved by updating these quantities for \(q=1,2,\dots,N\) in order.

Let’s formulate this carefully.

For \(q=1\), we have \(h_1=0, h_2=1, f_1=0, f_2=0\).

Hereafter, let \(\displaystyle p=\frac{q}{N}\).

First, focus on updating \(h\).
The probability that the player who plays right after the number of distinct integers has become \(q+1\) is different from the player who plays right after it has become \(q\) is expressed as \(\displaystyle (1-p) + (1-p)p^2 + (1-p)p^4 + \dots\).

Next, focus on updating \(f\).
The expected value of the fine increase for the player who is playing is expressed as \(p + p^3 + p^5 + \dots\), and for the player who is not playing, it is \(p^2 + p^4 + p^6 + \dots\).

Although it seems difficult to handle these infinite sums, by utilizing that \(\displaystyle 1 + p^2 + p^4 + p^6 + \dots = \sum_{i=0}^{\infty} p^{2i} = \frac{1}{1-p^2}\) holds for \(0 < p < 1\), all of them can be processed.

Implementing this calculation allows you to solve this problem in time complexity \(O(N)\).

Sample implementation (C++):

#include<bits/stdc++.h>

#define mod 998244353

long long power(long long a,long long b){
  long long x=1,y=a;
  while(b>0){
    if(b&1ll){
      x=(x*y)%mod;
    }
    y=(y*y)%mod;
    b>>=1;
  }
  return x%mod;
}

long long modular_inverse(long long n){
  return power(n,mod-2);
}

using namespace std;

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

  long long n;
  cin >> n;

  long long invn=modular_inverse(n);

  vector<long long> hand={0,1};
  vector<long long> fee={0,0};

  for(long long q=1;q<n;q++){
    long long prob=(q*invn)%mod;
    long long fce=modular_inverse(mod+1-((prob*prob)%mod));
    long long hce=((mod+1-prob)*fce)%mod;

    vector<long long> chand={
      (prob*hce)%mod,
      hce
    };
    vector<long long> cfee={
      (prob*fce)%mod,
      (((prob*prob)%mod)*fce)%mod,
    };

    vector<long long> nhand={0,0};

    for(long long i=0;i<2;i++){
      for(long long j=0;j<2;j++){
        nhand[(i+j)%2]+=hand[i]*chand[j];
        nhand[(i+j)%2]%=mod;

        fee[(i+j)%2]+=hand[i]*cfee[j];
        fee[(i+j)%2]%=mod;
      }
    }

    hand=nhand;
  }
  cout << fee[0]%mod << " " << fee[1]%mod << "\n";
  return 0;
}