//yukicoder@cpp17

#include <iostream>
#include <iomanip>

#include <algorithm>

#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>

#include <random>

#include <bitset>

#include <complex>

#include <utility>

#include <numeric>

#include <functional>


using namespace std;
using ll = long long;
using P = pair<ll,ll>;

const ll MOD = 998244353;
const ll MODx = 1000000007;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62LL)-1;
const double EPS = (1e-10);

P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};




template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }
 
 /*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/

/*
function corner below
*/


/*
Function corner above
*/

/* comment outed because can cause bugs
__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
*/

struct Combination {
  long long n_max;
  int mod;
  vector<long long> fac, inv, finv;
  Combination(int n, int mod):n_max(n), mod(mod), fac(n+1,1), inv(n+1,1), finv(n+1,1){
    for(int i = 2; n > i; i++){
      fac[i] = (fac[i-1]*i)%mod;
      inv[i] = mod-((inv[mod%i]*(mod/i))%mod);
      finv[i] = (finv[i-1]*inv[i])%mod;
    }
  }

  long long C(int n, int r){
    if(n < r)return 0;
    if(n > n_max)return 0;
    return (((fac[n]*finv[r])%mod)*finv[n-r])%mod;
  }

  long long P(int n, int r){
    if(n < r)return 0;
    if(n > n_max)return 0;
    return (fac[n]*finv[n-r])%mod;
  }

  long long H(int n, int r){
    if(n==0 && r==0)return 1;
    return C(n+r-1, r);
  }
};
int main(){
  int N;cin>>N;
  vector<int> A(2*N);
  set<int> B;
  for(int i = 0; 2*N > i; i++){
    cin>>A[i];
    if(i < N)B.insert(A[i]);
  }
  sort(A.begin(), A.end());
  int c = 0;
  for(int i = 0; N > i; i++){
    if(B.count(A[i]))c++;
  }
  Combination C(N+1, MOD);
  cout << (C.fac[c]*C.fac[N-c])%MOD << endl;
  return 0;
}