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;
int n;
int mod = 998244353;
long long fac[2123456], ifac[2123456];
long long powermod(long long a, long long b) {
  if (b == 0) return 1;
  long long t = powermod(a, b / 2);
  t = (t * t) % mod;
  if (b % 2 == 1) t = (t * a) % mod;
  return t;
}
void prepare() {
  fac[0] = 1;
  for (int i = 1; i <= 2123456 - 1; ++i) fac[i] = (fac[i - 1] * i) % mod;
  ifac[2123456 - 1] = powermod(fac[2123456 - 1], mod - 2);
  for (int i = 2123456 - 2; i >= 0; --i)
    ifac[i] = (ifac[i + 1] * (i + 1)) % mod;
}
long long cal(int n, int k) {
  long long rs = fac[n];
  rs = (rs * ifac[k]) % mod;
  rs = (rs * ifac[n - k]) % mod;
  return rs;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  prepare();
  long long rs = fac[n];
  for (int i = 1; i <= n - 1; ++i) {
    long long x = (cal(n, n - i) * fac[n - i]) % mod;
    x = (x * (fac[i] - 1 + mod) % mod) % mod;
    rs = (rs + x) % mod;
  }
  cout << rs;
  return 0;
}
