#3456. 城市规划

设 f n f_n fn​为 n n n个点的的点的简单无向连通图数目， g n g_n gn​为 n n n个点的简单无向图个数（不要求联通）。

对于 g n g_n gn​显然有 g n = 2 n ( n − 1 ) 2 g_n = 2 ^{\frac{n(n - 1)}{2}} gn​=22n(n−1)​，共有 n ( n + 1 ) 2 \frac{n(n + 1)}{2} 2n(n+1)​条边，然后每条边可选可不选。

我们枚举 1 1 1所在的点的连通块可得：
g n = ∑ i = 1 n C n − 1 i − 1 f i g n − i 2 ( 2 n ) = ∑ i = 1 n ( i − 1 n − 1 ) f i 2 ( 2 n − i ) 2 ( 2 n ) = ∑ i = 1 n ( n − 1 ) ! ( i − 1 ) ! ( n − i ) ! f i 2 ( 2 n − i ) 2 ( 2 n ) ( n − 1 ) ! = ∑ i = 1 n f i ( i − 1 ) ! 2 ( 2 n − i ) ( n − i ) ! 设 G ( x ) = ∑ n = 1 ∞ 2 ( 2 n ) ( n − 1 ) ! x n F ( x ) = ∑ n = 1 ∞ f n ( n − 1 ) ! H ( x ) = ∑ n = 0 ∞ 2 ( 2 n ) n ! G ( x ) = F ( x ) H ( x ) F ( x ) = G ( x ) H − 1 ( x ) 构 造 多 项 式 ， 多 项 式 求 逆 ， 把 [ x n ] 项 系 数 乘 上 ( n − 1 ) ! 即 是 答 案 g_n = \sum\limits_{i = 1} ^{n}C_{n - 1} ^{i - 1}f_ig_{n - i}\\ 2 ^{(_2 ^ n)} = \sum_{i = 1} ^{n}(_{i - 1} ^{n - 1})f_i 2 ^{(_2 ^{n - i})}\\ 2 ^{(_2 ^ n)} = \sum_{i = 1} ^{n} \frac{(n - 1)!}{(i - 1)!(n - i)!} f_i 2 ^{(_2 ^{n - i})}\\ \frac{2 ^{(_2 ^ n)}}{(n - 1)!} = \sum_{i = 1} ^{n} \frac{f_i}{(i - 1)!} \frac{2^{(_2 ^{n - i})}}{(n - i)!}\\ 设G(x) = \sum_{n = 1} ^{\infty} \frac{2 ^{(_2 ^n)}}{(n - 1)!} x ^ n\\ F(x) = \sum_{n = 1} ^{\infty} \frac{f_n}{(n - 1)!}\\ H(x) = \sum_{n = 0} ^{\infty} \frac{2 ^{(_2 ^n)}}{n!}\\ G(x) = F(x) H(x)\\ F(x) = G(x)H^{-1}(x)\\ 构造多项式，多项式求逆，把[x ^ n]项系数乘上(n - 1)!即是答案\\ gn​=i=1∑n​Cn−1i−1​fi​gn−i​2(2n​)=i=1∑n​(i−1n−1​)fi​2(2n−i​)2(2n​)=i=1∑n​(i−1)!(n−i)!(n−1)!​fi​2(2n−i​)(n−1)!2(2n​)​=i=1∑n​(i−1)!fi​​(n−i)!2(2n−i​)​设G(x)=n=1∑∞​(n−1)!2(2n​)​xnF(x)=n=1∑∞​(n−1)!fn​​H(x)=n=0∑∞​n!2(2n​)​G(x)=F(x)H(x)F(x)=G(x)H−1(x)构造多项式，多项式求逆，把[xn]项系数乘上(n−1)!即是答案

#include <bits/stdc++.h>

using namespace std;

const int mod = 1004535809, inv2 = mod + 1 >> 1;

namespace Quadratic_residue {
  struct Complex {
    int r, i;

    Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}
  };

  int I2;

  Complex operator * (const Complex &a, Complex &b) {
    return Complex((1ll * a.r * b.r % mod  + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);
  }

  Complex quick_pow(Complex a, int n) {
    Complex ans = Complex(1, 0);
    while (n) {
      if (n & 1) {
        ans = ans * a;
      }
      a = a * a;
      n >>= 1;
    }
    return ans;
  }

  int get_residue(int n) {
    mt19937 e(233);
    if (n == 0) {
      return 0;
    }
    if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {
      return -1;
    }
    uniform_int_distribution<int> r(0, mod - 1);
    int a = r(e);
    while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {
      a = r(e);
    }
    I2 = (1ll * a * a % mod - n + mod) % mod;
    int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;
    if(x > y) swap(x, y);
    return x;
  }
}

const int N = 1e6 + 10;

int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];

int quick_pow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans = 1ll * a * ans % mod;
    }
    a = 1ll * a * a % mod;
    n >>= 1;
  }
  return ans;
}

void get_r(int lim) {
  for (int i = 0; i < lim; i++) {
    r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
  }
}

void get_inv(int n) {
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
  }
}

void NTT(int *f, int lim, int rev) {
  for (int i = 0; i < lim; i++) {
    if (i < r[i]) {
      swap(f[i], f[r[i]]);
    }
  }
  for (int mid = 1; mid < lim; mid <<= 1) {
    int wn = quick_pow(3, (mod - 1) / (mid << 1));
    for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
        int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
        f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
      }
    }
  }
  if (rev == -1) {
    int inv = quick_pow(lim, mod - 2);
    reverse(f + 1, f + lim);
    for (int i = 0; i < lim; i++) {
      f[i] = 1ll * f[i] * inv % mod;
    }
  }
}

void polyinv(int *f, int *g, int n) {
  if (n == 1) {
    g[0] = quick_pow(f[0], mod - 2);
    return ;
  }
  polyinv(f, g, n + 1 >> 1);
  for (int i = 0; i < n; i++) {
    t[i] = f[i];
  }
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  get_r(lim);
  NTT(t, lim, 1);
  NTT(g, lim, 1);
  for (int i = 0; i < lim; i++) {
    int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;
    g[i] = 1ll * g[i] * cur % mod;
    t[i] = 0;
  }
  NTT(g, lim, -1);
  for (int i = n; i < lim; i++) {
    g[i] = 0;
  }
}

void polysqrt(int *f, int *g, int n) {
  if (n == 1) {
    g[0] = Quadratic_residue::get_residue(f[0]);
    return ;
  }
  polysqrt(f, g, n + 1 >> 1);
  polyinv(g, b, n);
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  get_r(lim);
  for (int i = 0; i < n; i++) {
    t[i] = f[i];
  }
  NTT(g, lim, 1);
  NTT(b, lim, 1);
  NTT(t, lim, 1);
  for (int i = 0; i < lim; i++) {
    g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;
    b[i] = t[i] = 0;
  }
  NTT(g, lim, -1);
  for (int i = n; i < lim; i++) {
    g[i] = 0;
  }
}

void derivative(int *a, int *b, int n) {
  for (int i = 0; i < n; i++) {
    b[i] = 1ll * a[i + 1] * (i + 1) % mod;
  }
}

void integrate(int *a, int n) {
  for (int i = n - 1; i >= 1; i--) {
    a[i] = 1ll * a[i - 1] * inv[i] % mod;
  }
  a[0] = 0;
}

void polyln(int *f, int *g, int n) {
  polyinv(f, b, n);
  derivative(f, g, n);
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  get_r(lim);
  NTT(g, lim, 1);
  NTT(b, lim, 1);
  for (int i = 0; i < lim; i++) {
    g[i] = 1ll * g[i] * b[i] % mod;
    b[i] = 0;
  }
  NTT(g, lim, -1);
  for (int i = n; i < lim; i++) {
    g[i] = 0;
  }
  integrate(g, n);
}

void polyexp(int *f, int *g, int n) {
  if (n == 1) {
    g[0] = 1;
    return ;
  }
  polyexp(f, g, n + 1 >> 1);
  int lim = 1;
  while (lim < 2 * n) {
    lim <<= 1;
  }
  polyln(g, d, n);
  for (int i = 0; i < n; i++) {
    t[i] = (f[i] - d[i] + mod) % mod;
  }
  t[0] = (t[0] + 1) % mod;
  get_r(lim);
  NTT(g, lim, 1);
  NTT(t, lim, 1);
  for (int i = 0; i < lim; i++) {
    g[i] = 1ll * g[i] * t[i] % mod;
    t[i] = d[i] =  0;
  }
  NTT(g, lim, -1);
  for (int i = n; i < lim; i++) {
    g[i] = 0;
  }
}

/*
  b存放多项式逆，
  c存放多项式开根，
  d存放多项式对数ln，
  e存放多项式指数exp，
  t作为中间转移数组,
  如果要用到polyln，得提前调用get_inv(n)先预先得到我们想要得到的逆元范围。
*/

int h[N], g[N], fac[N], ifac[N], n;

void init() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = 1ll * i * fac[i - 1] % mod;
  }
  ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) {
    ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  }
}

int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; i++) {
    g[i] = 1ll * quick_pow(2, 1ll * i * (i - 1) / 2 % (mod - 1)) * ifac[i - 1] % mod;
  }
  for (int i = 0; i <= n; i++) {
    h[i] = 1ll * quick_pow(2, 1ll * i * (i - 1) / 2 % (mod - 1)) * ifac[i] % mod;
  }
  polyinv(h, b, n + 1);
  for (int i = 0; i <= n; i++) {
    h[i] = b[i];
    b[i] = 0;
  }
  int lim = 1;
  while (lim <= 2 * n) {
    lim <<= 1;
  }
  get_r(lim);
  NTT(g, lim, 1);
  NTT(h, lim, 1);
  for (int i = 0; i < lim; i++) {
    g[i] = 1ll * g[i] * h[i] % mod;
    h[i] = 0;
  }
  NTT(g, lim, -1);
  printf("%d\n", 1ll * g[n] * fac[n - 1] % mod);
  return 0;
}