题意

略。（语文太差）

题解

首先一个结论：随机一个\([0, 1]\)之间的实数序列，只用到各个位置相互之间的大小关系，每种关系出现的概率等同于随机一个\([1, n]\)的排列。
原因是出现某些位置值相同的概率是无穷小，并且只有有限种情况出现相同的值，因此可以忽略。
然后在草稿纸上画一画，就知道如果\(p_1, p_2, \ldots, p_x \in [n - x + 1, n]\)，那么\(1, 2, \ldots, x\)和\(x + 1, x + 2, \ldots, n\)之间必然没有边。
考虑dp。但是如果直接dp答案会有点难，所以设\(f_n\)表示长度为\(n\)的序列，不存在一个\(x\)，使得\(1, 2, \ldots, x\)和\(x + 1, x + 2, \ldots, n\)没有边的不同序列数，并规定\(f_0 = 1\)。
考虑补集转化，有
\[ f_n = n! - \sum_{i = 1} ^ {n - 1} f_{n - i} i! \\ 2n! = \sum_{i = 0} ^ {n} f_{n - i} i! + [n = 0] \\ \]
考虑其生成函数
\[ F(x) = \sum_{i \geq 0} f_i x ^ i \]
与一个辅助生成函数
\[ G(x) = \sum_{i \geq 0} i! x ^ i \]
则
\[ 2G(x) = G(x)F(x) + 1 \\ F(x) = \frac{2G(x) - 1}{G(x)} \\ \]
这可以用多项式求逆求出。
考虑答案。设\(H(x) = \sum_{i \geq 0} i f_i x ^ i\)（此时\(f_i\)已求出），则有
\[ ans = [x ^ n] \sum_{k \geq 0} H(X) ^ k = [x ^ n] \frac{1}{1 - H(x)} \]
再做一次多项式求逆即可。
复杂度\(\mathcal O(n \log n)\)。

#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector <int> poly;
const int N = 1 << 20, mod = 998244353, G = 3;
const ll infty = 1ll * mod * mod;
int power (int a, int b) {
    int ret = 1;
    if (b < 0) {
        b += mod - 1;
    }
    for ( ; b; b >>= 1, a = 1ll * a * a % mod) {
        if (b & 1) {
            ret = 1ll * ret * a % mod;
        }
    }
    return ret;
}
namespace {
    void printp (poly a) {
        if (!a.size()) {
            return;
        }
        printf("%d", a[0]);
        for (int i = 1; i < (int)a.size(); ++i) {
            printf(" %d", a[i]);
        }
        putchar('\n');
    }
    int adjust (int n) {
        int ret = 1;
        for ( ; ret < n; ret <<= 1);
        return ret;
    }
    poly trans (int n, int *a) {
        int m = adjust(n);
        poly ret; ret.resize(m, 0);
        for (int i = 0; i < n; ++i) {
            ret[i] = a[i];
        }
        return ret;
    }
    void dnt (int n, poly &_a) {
        static int rev[N], a[N], wi[N];
        for (int i = 0; i < n; ++i) {
            rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? n >> 1 : 0);
            a[i] = _a[rev[i]];
        }
        for (int l = 2, _w; l <= n; l <<= 1) {
            _w = power(G, (mod - 1) / l), wi[l >> 1] = 1;
            for (int i = (l >> 1) + 1; i < l; ++i) {
                wi[i] = 1ll * wi[i - 1] * _w % mod;
            }
        }
        for (int l = 2, m = 1; l <= n; m = l, l <<= 1) {
            for (int i = 0; i < n; i += l) {
                int *u = a + i, *v = a + i + m, *w = wi + m;
                for (int j = 0, x, y; j < m; ++u, ++v, ++w, ++j) {
                    x = *u, y = 1ll * (*v) * (*w) % mod;
                    *u = (x + y) % mod, *v = (x - y + mod) % mod;
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            _a[i] = a[i];
        }
    }
    void idnt (int n, poly &_a) {
        static int rev[N], a[N], wi[N];
        for (int i = 0; i < n; ++i) {
            rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? n >> 1 : 0);
            a[i] = _a[rev[i]];
        }
        for (int l = 2, _w; l <= n; l <<= 1) {
            _w = power(G, mod - 1 - (mod - 1) / l), wi[l >> 1] = 1;
            for (int i = (l >> 1) + 1; i < l; ++i) {
                wi[i] = 1ll * wi[i - 1] * _w % mod;
            }
        }
        for (int l = 2, m = 1; l <= n; m = l, l <<= 1) {
            for (int i = 0; i < n; i += l) {
                int *u = a + i, *v = a + i + m, *w = wi + m;
                for (int j = 0, x, y; j < m; ++u, ++v, ++w, ++j) {
                    x = *u, y = 1ll * (*v) * (*w) % mod;
                    *u = (x + y) % mod, *v = (x - y + mod) % mod;
                }
            }
        }
        int invn = power(n, mod - 2);
        for (int i = 0; i < n; ++i) {
            _a[i] = 1ll * a[i] * invn % mod;
        }
    }
    poly conv (int n, poly a, poly b, int f = 0) {
        a.resize(n, 0), b.resize(n, 0);
        n <<= 1, a.resize(n, 0), b.resize(n, 0);
        dnt(n, a), dnt(n, b);
        for (int i = 0; i < n; ++i) {
            a[i] = 1ll * a[i] * b[i] % mod;
            if (f) {
                a[i] = 1ll * a[i] * b[i] % mod;
            }
        }
        idnt(n, a);
        return a;
    }
    poly plu (int n, poly a, poly b) {
        for (int i = 0; i < n; ++i) {
            if ((a[i] += b[i]) >= mod) {
                a[i] -= mod;
            }
        }
        return a;
    }
    poly minus (int n, poly a, poly b) {
        for (int i = 0; i < n; ++i) {
            if ((a[i] -= b[i]) < 0) {
                a[i] += mod;
            }
        }
        return a;
    }
    poly inv (int n, poly f) {
        if (n == 1) {
            return (poly) {power(f[0], mod - 2)};
        }
        f.resize(n);
        poly g = inv(n >> 1, f);
        g.resize(n, 0);
        f = conv(n, f, g, 1);
        g = minus(n, plu(n, g, g), f);
        return g.resize(n), g;
    }
}
int fac[N], c[N];
poly A, B, C;
int solve (int n) {
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) {
        fac[i] = 1ll * fac[i - 1] * i % mod;
    }
    A = trans(n, fac);
    B = plu(A.size(), A, A), B[0] = (B[0] - 1 + mod) % mod;
    A = inv(A.size(), A);
    B = conv(B.size(), B, A);
    for (int i = 0; i < n; ++i) {
        c[i] = (mod - 1ll * i * B[i] % mod) % mod;
    }
    c[0] = (c[0] + 1) % mod;
    C = trans(n, c);
    C = inv(C.size(), C);
    return C[n - 1];
}
int n;
int main () {
    cin >> n;
    cout << solve(n + 1) << endl;
    return 0;
}