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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
string direc = "URDL";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int fac[1000035], inv[1000035];
void initfac(int n) {
  fac[0] = inv[0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    fac[i] = (long long)fac[i - 1] * i % 998244353;
    inv[i] = fastPow(fac[i], 998244353 - 2);
  }
}
int C(int n, int m) {
  if (n < m) return 0;
  return (long long)fac[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
}
long long c[1000035];
void fmain(int tid) {
  scanf("%d", &n);
  initfac(n);
  long long S = (long long)n * fac[n] % 998244353;
  int ans = S * (S + 1) % 998244353 * fastPow(2, 998244353 - 2) % 998244353;
  int ss = 0;
  c[0] = n;
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    long long a = (long long)fac[n] * inv[n - i + 1] % 998244353;
    long long b = (long long)(n - i) * (n - i) % 998244353 + 1;
    c[i] = a * b % 998244353;
    addmod(ss, c[i] * i % 998244353);
  }
  for (int i = n; i < n + n; i++) {
    addmod(ss, (long long)(fac[n] - c[i - n] + 998244353) % 998244353 * i %
                   998244353);
  }
  addmod(ans, 998244353 - ss);
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
