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>
#pragma GCC optimize "trapv"
const int inf1 = 1e9;
const long long int inf2 = 1e18;
const int N = 100000;
using namespace std;
int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
long long int fact[10 * N + 1], infact[10 * N + 1];
long long int xymodp(long long int first, long long int second,
                     long long int p) {
  long long int a = 1;
  first %= p;
  while (second) {
    if (second & 1) a = (a * first) % p;
    first = (first * first) % p;
    second /= 2;
  }
  return a;
}
void init() {
  fact[0] = 1;
  for (int i = 1; i < 1000001; i++) fact[i] = (i * fact[i - 1]) % 998244353;
  for (int i = 1; i < 1000001; i++)
    infact[i] = xymodp(fact[i], 998244353 - 2, 998244353);
  infact[0] = 1;
  infact[1] = 1;
}
long long int cat_num(long long int n, long long int m) {
  long long int k = fact[n + m];
  k *= infact[n];
  k %= 998244353;
  k *= infact[m];
  k %= 998244353;
  k *= (n - m + 1);
  k %= 998244353;
  k *= xymodp(n + 1, 998244353 - 2, 998244353);
  return k % 998244353;
}
int solve() {
  long long int n;
  cin >> n;
  long long int ans = n * fact[n] % 998244353;
  long long int cnt = 0;
  for (int i = 0; i < n - 1; i++) {
    cnt = (cnt + (fact[n] * infact[i + 1]) % 998244353) % 998244353;
  }
  ans = ans - cnt;
  ans %= 998244353;
  ans = (ans + 998244353) % 998244353;
  cout << ans << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
