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;
using namespace std;
const int maxn = 1e6 + 100;
long long mod = 998244353;
long long f[maxn];
long long n;
long long ans;
long long tav(long long i, long long j) {
  if (j == 0) return 1;
  if (j == 1) return (i % mod);
  long long s = tav(i, j / 2);
  s = (1LL * s * s) % mod;
  if (j % 2) s = (s * i) % mod;
  return s;
}
long long ent(long long i, long long j) {
  long long z = f[j];
  z = (z * tav(f[i], mod - 2)) % mod;
  z = (z * tav(f[j - i], mod - 2)) % mod;
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  f[0] = 1;
  for (long long y = 1; y <= n; y++) f[y] = (f[y - 1] * y) % mod;
  for (long long y = 1; y < n; y++) {
    long long z = ent(y, n);
    z = z = (((z * (f[y] - 1)) % mod) + mod) % mod;
    z = (z * f[n - y]) % mod;
    ans = (ans + z) % mod;
  }
  ans = (ans + f[n]) % mod;
  cout << ans;
}
