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;
const int MOD = 998244353;
struct mod_int {
  int val;
  mod_int(long long v = 0) {
    if (v < 0) v = v % MOD + MOD;
    if (v >= MOD) v %= MOD;
    val = v;
  }
  static int mod_inv(int a, int m = MOD) {
    int g = m, r = a, x = 0, y = 1;
    while (r != 0) {
      int q = g / r;
      g %= r;
      swap(g, r);
      x -= q * y;
      swap(x, y);
    }
    return x < 0 ? x + m : x;
  }
  explicit operator int() const { return val; }
  mod_int &operator+=(const mod_int &other) {
    val += other.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  mod_int &operator-=(const mod_int &other) {
    val -= other.val;
    if (val < 0) val += MOD;
    return *this;
  }
  static unsigned fast_mod(uint64_t x, unsigned m = MOD) {
    return x % m;
    unsigned x_high = x >> 32, x_low = (unsigned)x;
    unsigned quot, rem;
    asm("divl %4\n" : "=a"(quot), "=d"(rem) : "d"(x_high), "a"(x_low), "r"(m));
    return rem;
  }
  mod_int &operator*=(const mod_int &other) {
    val = fast_mod((uint64_t)val * other.val);
    return *this;
  }
  mod_int &operator/=(const mod_int &other) { return *this *= other.inv(); }
  friend mod_int operator+(const mod_int &a, const mod_int &b) {
    return mod_int(a) += b;
  }
  friend mod_int operator-(const mod_int &a, const mod_int &b) {
    return mod_int(a) -= b;
  }
  friend mod_int operator*(const mod_int &a, const mod_int &b) {
    return mod_int(a) *= b;
  }
  friend mod_int operator/(const mod_int &a, const mod_int &b) {
    return mod_int(a) /= b;
  }
  mod_int &operator++() {
    val = val == MOD - 1 ? 0 : val + 1;
    return *this;
  }
  mod_int &operator--() {
    val = val == 0 ? MOD - 1 : val - 1;
    return *this;
  }
  mod_int operator++(int) {
    mod_int before = *this;
    ++*this;
    return before;
  }
  mod_int operator--(int) {
    mod_int before = *this;
    --*this;
    return before;
  }
  mod_int operator-() const { return val == 0 ? 0 : MOD - val; }
  bool operator==(const mod_int &other) const { return val == other.val; }
  bool operator!=(const mod_int &other) const { return val != other.val; }
  mod_int inv() const { return mod_inv(val); }
  mod_int pow(long long p) const {
    assert(p >= 0);
    mod_int a = *this, result = 1;
    while (p > 0) {
      if (p & 1) result *= a;
      a *= a;
      p >>= 1;
    }
    return result;
  }
  friend ostream &operator<<(ostream &stream, const mod_int &m) {
    return stream << m.val;
  }
};
vector<mod_int> inv = {0, 1}, factorial = {1, 1}, inv_factorial = {1, 1};
void prepare_factorials(int maximum) {
  for (int p = 2; p * p <= MOD; p++) assert(MOD % p != 0);
  inv.resize(maximum + 1);
  factorial.resize(maximum + 1);
  inv_factorial.resize(maximum + 1);
  for (int i = 2; i <= maximum; i++) {
    inv[i] = inv[MOD % i] * (MOD - MOD / i);
    factorial[i] = i * factorial[i - 1];
    inv_factorial[i] = inv[i] * inv_factorial[i - 1];
  }
}
mod_int choose(long long n, long long r) {
  if (r < 0 || r > n) return 0;
  return factorial[n] * inv_factorial[r] * inv_factorial[n - r];
}
mod_int inv_choose(long long n, long long r) {
  assert(0 <= r && r <= n);
  return inv_factorial[n] * factorial[r] * factorial[n - r];
}
mod_int permute(long long n, long long k) {
  if (k < 0 || k > n) return 0;
  return factorial[n] * inv_factorial[n - k];
}
mod_int inv_permute(long long n, long long k) {
  assert(0 <= k && k <= n);
  return inv_factorial[n] * factorial[n - k];
}
int N;
mod_int get_freq(int f) {
  if (f == 0) return N;
  if (f < N) {
    mod_int freq = mod_int(N - f) * (N - f) + 1;
    freq *= permute(N, f - 1);
    return freq;
  }
  return factorial[N] - get_freq(f - N);
};
int main() {
  cin >> N;
  prepare_factorials(2 * N);
  mod_int total = 0;
  for (int f = 0; f < 2 * N; f++) total += f * get_freq(f);
  mod_int PN = N * factorial[N];
  total = PN * (PN + 1) / 2 - total;
  cout << total << '\n';
}