#include <bits/stdc++.h>
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define Yes cout << "Yes" << "\n"
#define No cout << "No" << "\n"
#define rtr0 return(0)
#define all(x) x.begin(), x.end()
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using mint=static_modint<998244353>;
//using mint=static_modint<1000000007>;
////using mint=modint;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

using ll=long long;
using l3=__int128;
using ull=unsigned long long;
using ld=long double;
using P=pair<ll,ll>;
const ld PI=acos(-1);
template<typename T>bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<typename T>bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
void yn(bool f){cout<<(f?"Yes":"No")<<endl;}
const vector<int> dx={1,0,-1,0};
const vector<int> dy={0,1,0,-1};
const int inf=1001001001;
const ll INF=1001001001001001001;

ll mod=998244353;

//繰り返し二乗法　n^k modはグローバルに
ll pow_mod(ll n,ll k){
    if(k==0)return 1;
    if(k%2==1)return pow_mod(n,k-1)*n%mod;
    ll t=pow_mod(n,k/2);
    return t*t%mod;
}


void solve(){
    ll N;cin>>N;
    mint n=N;
    mint ans=(2-n+n*(n-1)/2);
    mint p=1;
    rep(i,4)p*=(n-i);
    p/=24;
    p*=25;
    p/=36;
    cout<<(ans+p).val()<<endl;
}

int main(){
    ll t=1;
    //cin>>t;
    rep(i,t)solve();
}