Let n be an integer. Consider all permutations on integers 1 to n in lexicographic order, and concatenate them into one big sequence P. For example, if n = 3, then P = [1, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 3, 1, 2, 3, 2, 1]. The length of this sequence is n ⋅ n!.

Let 1 ≤ i ≤ j ≤ n ⋅ n! be a pair of indices. We call the sequence (P_i, P_{i+1}, ..., P_{j-1}, P_j) a subarray of P. 

You are given n. Find the number of distinct subarrays of P. Since this number may be large, output it modulo 998244353 (a prime number). 

Input

The only line contains one integer n (1 ≤ n ≤ 10^6), as described in the problem statement.

Output

Output a single integer — the number of distinct subarrays, modulo 998244353.

Examples

Input


2


Output


8


Input


10


Output


19210869

Note

In the first example, the sequence P = [1, 2, 2, 1]. It has eight distinct subarrays: [1], [2], [1, 2], [2, 1], [2, 2], [1, 2, 2], [2, 2, 1] and [1, 2, 2, 1]. 
#include <bits/stdc++.h>
using namespace std;
inline void readI(int *i) {
  register int t = 0;
  register char z = getchar();
  int znak = 1;
  if (z == '-') {
    znak = -1;
  }
  while ((z < '0') || ('9' < z)) {
    z = getchar();
    if (z == '-') {
      znak = -1;
    }
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *i = (t * znak);
}
inline void readUI(int *i) {
  register int t = 0;
  register char z = getchar();
  while ((z < '0') || ('9' < z)) {
    z = getchar();
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *i = t;
}
inline void readL(long long int *l) {
  register long long int t = 0;
  register char z = getchar();
  int znak = 1;
  if (z == '-') {
    znak *= -1;
  }
  while ((z < '0') || ('9' < z)) {
    z = getchar();
    if (z == '-') {
      znak = -1;
    }
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *l = (t * znak);
}
inline void readUL(long long int *l) {
  register long long int t = 0;
  register char z = getchar();
  while ((z < '0') || ('9' < z)) {
    z = getchar();
  }
  while (('0' <= z) && (z <= '9')) {
    t = (t << 3) + (t << 1) + z - '0';
    z = getchar();
  }
  *l = t;
}
inline void writeL(long long int l) {
  if (l == 0) {
    putchar(48);
  } else {
    if (l < 0) {
      l *= -1;
      putchar(45);
    }
    int _tab[21];
    int wsk = 0;
    while (l > 0) {
      ++wsk;
      _tab[wsk] = (l % 10) + 48;
      l /= 10;
    }
    for (int j = wsk; j >= 1; --j) {
      putchar(_tab[j]);
    }
  }
}
inline void writeS(string s) {
  int l = s.length();
  for (int i = 0; i < l; ++i) {
    putchar(s[i]);
  }
}
inline void space() { putchar(32); }
inline void endl() { putchar(10); }
int n;
long long int silnia[(int)(1e6) + 5];
long long int P = 998244353;
long long int wynik, ile;
long long int pot(long long int a, int b) {
  if (!b) {
    return 1;
  }
  if (!(b & 1)) {
    long long int s = pot(a, b >> 1);
    s *= s;
    return s % P;
  }
  long long int s = pot(a, b - 1);
  s *= a;
  return s % P;
}
long long int dziel(long long int war, long long int a, long long int b) {
  war *= a;
  war %= P;
  war *= pot(b, P - 2);
  war %= P;
  return war;
}
int main() {
  readUI(&n);
  silnia[0] = 1;
  for (int i = 1; i <= n; ++i) {
    silnia[i] = silnia[i - 1] * i;
    silnia[i] %= P;
  }
  wynik = silnia[n];
  ile = silnia[n];
  for (int i = 2; i <= n; ++i) {
    long long int pom = dziel(ile, i - 1, i);
    wynik += pom * (n - i);
    ile -= pom;
    wynik %= P;
    ile += P;
    ile %= P;
  }
  cout << wynik;
  return 0;
}
