#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], al;
int b[N], bl;

void mul(int a[], int &al, int b[], int &bl) {
  int t = 0;
  int c[N] = {0};
  for (int i = 0; i < al; i++)
    for (int j = 0; j < bl; j++)
      c[i + j] += a[i] * b[j];
  al += bl;
  for (int i = 0; i < al; i++) {
    t += c[i];
    a[i] = t % 10;
    t /= 10;
  }
  while (al > 0 && a[al - 1] == 0)
    al--;
}
void jiecheng(int x) {
  a[0] = 1;
  al = 1;

  for (int i = 2; i <= x; i++) {
    int t = i;
    bl = 0;
    while (t > 0) {
      b[bl++] = t % 10;
      t /= 10;
    }
    mul(a, al, b, bl);
  }
}

void add(int a[], int &al, int b[], int &bl) {
  int t = 0;
  al = max(al, bl);
  for (int i = 0; i < al; i++) {
    t += a[i] + b[i];
    a[i] = t % 10;
    t /= 10;
  }
  if (t)
    a[al++] = 1;
  while (al > 0 && a[al - 1] == 0)
    al--;
}

int main() {
  int n;
  cin >> n; // 读入n

  int sum[N] = {0}; // 存储总和
  int suml = 0;     // 总和的长度

  for (int i = 1; i <= n; i++) {
    jiecheng(i);           // 计算i的阶乘，结果存在a数组中
    add(sum, suml, a, al); // 将a数组的结果加到sum数组中
  }

  // 从高位到低位输出结果
  for (int i = suml - 1; i >= 0; i--) {
    cout << sum[i];
  }
  cout << endl;

  return 0;
}