#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <random>
#include <climits>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <iomanip>
 
using namespace std;
 
#define endl '\n'
typedef long long ll;
#define int long long
typedef __int128_t int128;
typedef pair<int, int> PII;
typedef pair<PII, int> PIII;
const int INF = 0x3f3f3f3f, mod = 998244353;
mt19937 rng(time(0));
#define x first
#define y second
#define lowbit(x) ((x) & -(x))
//#define double long double
const int N = 10000010;
const int M = 10000000;
ll n, m;
ll fact[N], infact[N];

int qmi(int a, int k, int p)
{
    int res = 1;
    while(k)
    {
        if(k & 1) res = (ll)res * a % p;
        k >>= 1;
        a = (ll)a * a % p;
    }
    return res;
}

ll s[N]; //n个数的前缀积
ll sv[N]; //n个数的前缀积的逆元
ll inv[N]; //第i个数的逆元



void solve() {
    
    cin >> n;
    ll ans = 0;
    
    for (int i = 1; i <= n; i ++ ) {
        if (n - i - (i - 1) >= 0) {
            ans = (ans + fact[n - i] * infact[n - i - (i - 1)] % mod * fact[n - i] % mod) % mod;
        }
        
    }
    cout << ans << endl;
}
 
signed main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//     freopen("input.txt", "r", stdin);
//      freopen("output.txt", "w", stdout);
    int t = 1;
    //cin >> t;
    
    s[0] = 1;
    for (int i = 1; i <= M; ++i) s[i] = s[i - 1] * i % mod;
    sv[M] = qmi(s[M], mod - 2, mod);
    // 当然这里也可以用 exgcd 来求逆元
    for (int i = M; i >= 1; --i) sv[i - 1] = sv[i] * i % mod;
    for (int i = 1; i <= M; ++i) inv[i] = sv[i] * s[i - 1] % mod;
    

    fact[0] = infact[0] = 1;
    for (int i = 1; i <= M; i ++ )
    {
        fact[i] = (ll)i * fact[i - 1] % mod;
        infact[i] = (ll)infact[i - 1] * inv[i] % mod;
    }
    while(t--)
    {
        solve();
    }
      
    return 0;
}