#include <bits/stdc++.h>
using namespace std;
#define fs first
#define endl '\n'
#define sc second
#define pb push_back
#define int long long
#define bit(x) (1ll << (x))
#define sz(x) (int)(x).size()
#define al(x) (x).begin(),(x).end()
#define all(x) (x).begin() + 1,(x).end()
using LL = long long;
using VI = vector <int>;
using VS = vector <string>;
using PII = pair <int, int>;
using I64 = unsigned long long;
using VVI = vector < vector<int> >;
template<LL N> using AL = array<LL, N>;
template<int N> using AI = array<int, N>;
const int inf = bit(60), mod = 998244353, N = 1e7 + 6;

int fpow(int x, int t = mod - 2) {
    int res = 1;
    for (; t; t >>= 1, x = 1ll * x * x % mod)
        if (t & 1) res = 1ll * res * x % mod;
    return res;
}

int fac[N], ifac[N], deg[N], ans[N];

int C(int n, int m) {
    if (n < m) return 0;
    return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}

void init() //这里是处理阶乘和每个数字的逆元，fac[i]存的是i!,ifac[i]存的是i!的逆元，方便运算
{
    fac[0] = ifac[0] = 1;
    for (int i = 1; i < N; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[N - 1] = fpow(fac[N - 1]);
    for (int i = N - 2; i; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}

void solve() {
    init();
    int n; cin >> n;
    int res = 0;
    for(int i = 1;i * 2 - 1 <= n;i++) {
        res = (res + C(n - i, i - 1) * fac[i - 1] % mod * fac[n - i] % mod) % mod;
    }
    cout << res << endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed << setprecision(10);
    int T = 1;
//    cin >> T;
    while(T--) solve();

    return 0;
}