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 mod = 998244353;
long long bin_pow(long long x, long long pw) {
  if (pw == 1) return x % mod;
  long long y = bin_pow(x, pw / 2);
  if (pw % 2 == 1)
    return (((x * y) % mod) * y) % mod;
  else
    return (y * y) % mod;
}
long long rev(long long x) { return bin_pow(x, mod - 1); }
int main() {
  long long n = 0;
  cin >> n;
  vector<long long> fact(n + 1), C(n + 1);
  fact[0] = 1;
  C[0] = 1;
  for (long long i = 1; i <= n; ++i) {
    fact[i] = fact[i - 1] * i % mod;
    C[i] = C[i - 1] * (n - i + 1) * rev(i) % mod;
  }
  long long ans = fact[n];
  for (int k = 1; k < n; ++k)
    ans = (ans + C[k] * (fact[n - k] - 1) % mod) % mod;
  cout << ans << '\n';
}
