Problem: 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]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long max(long long x, long long y) { return x > y ? x : y; }
inline long long min(long long x, long long y) { return x < y ? x : y; }
inline long long gcd(long long n, long long m) { return m ? gcd(m, n % m) : n; }
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
inline void swap(long long &x, long long &y) {
  long long t = x;
  x = y;
  y = t;
  return;
}
namespace yspm {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
const long long mod = 998244353, N = 2e6 + 10;
inline long long add(long long x, long long y) {
  return x + y >= mod ? x + y - mod : x + y;
}
inline long long del(long long x, long long y) {
  return x - y < 0 ? x - y + mod : x - y;
}
inline long long mul(long long x, long long y) {
  return x * y - x * y / mod * mod;
}
inline long long ksm(long long x, long long y) {
  long long res = 1;
  for (; y; y >>= 1, x = mul(x, x))
    if (y & 1) res = mul(res, x);
  return res;
}
long long ans, fac[N], inv[N], ifac[N], c[N], n;
signed main() {
  n = read();
  inv[0] = inv[1] = ifac[0] = ifac[1] = fac[0] = fac[1] = 1;
  for (register long long i = 2; i <= N - 1; ++i)
    fac[i] = mul(fac[i - 1], i), inv[i] = mod - mul(mod / i, inv[mod % i]),
    ifac[i] = mul(ifac[i - 1], inv[i]);
  ans = mul(inv[2], mul(mul(n, fac[n]), add(mul(n, fac[n]), 1)));
  c[0] = n;
  for (register long long i = 1; i <= n - 1; ++i)
    ans = del(ans, mul(i, c[i] = mul(fac[n], mul(ifac[n - i + 1],
                                                 add(1, mul(n - i, n - i))))));
  for (register long long i = n; i <= 2 * n - 1; ++i)
    ans = del(ans, mul(i, del(fac[n], c[i - n])));
  cout << ans << endl;
  return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }