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 will be 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. Its length is defined as the number of its elements, i.e., j - i + 1. Its sum is the sum of all its elements, i.e., ∑_{k=i}^j p_k. 

You are given n. Find the number of subarrays of p of length n having sum (n(n+1))/(2). 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 subarrays of length n having sum (n(n+1))/(2), modulo 998244353.

Examples

Input


3


Output


9


Input


4


Output


56


Input


10


Output


30052700

Note

In the first sample, there are 16 subarrays of length 3. In order of appearance, they are:

[1, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 2], [3, 2, 2], [2, 2, 1], [2, 1, 3], [1, 3, 2], [3, 2, 3], [2, 3, 1], [3, 1, 3], [1, 3, 1], [3, 1, 2], [1, 2, 3], [2, 3, 2], [3, 2, 1]. 

Their sums are 6, 6, 7, 6, 7, 5, 6, 6, 8, 6, 7, 5, 6, 6, 7, 6. As (n(n+1))/(2) = 6, the answer is 9.
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) { return a + b - max(a, b); }
long long mod(long long a, long long b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long ans(long long i, long long n) {
  long long terms = (n / i);
  return (terms * (2 + (terms - 1) * i)) / 2;
}
long long mulmod(long long a, long long b, long long mod) {
  long long res = 0;
  a = a % mod;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % mod;
    a = (a * 2) % mod;
    b /= 2;
  }
  return res % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long mod = 998244353;
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return 0;
  }
  long long ans = 0;
  long long cur = n;
  long long ctr = n - 1;
  long long sum = 0;
  for (long long i = 1; i <= n - 1; i++) {
    ans += mulmod(i, (cur - sum + mod) % mod, mod);
    ans %= mod;
    sum = cur % mod;
    cur = mulmod(cur, ctr, mod);
    ctr--;
  }
  cout << ans % mod;
}
