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;
vector<long long int> fact(1000007);
inline long long int mdpower(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res % 998244353 * a % 998244353) % 998244353;
    b /= 2;
    a = (a % 998244353 * a % 998244353) % 998244353;
  }
  return res;
}
inline long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    b /= 2;
    a = a * a;
  }
  return res;
}
inline void factorial() {
  fact[0] = 1;
  for (long long int i = 1; i < 1000007; i++)
    fact[i] = (fact[i - 1] % 998244353 * i % 998244353) % 998244353;
}
inline long long int inverse(long long int a) {
  return mdpower(a, 998244353 - 2);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  factorial();
  long long int n;
  cin >> n;
  long long int ans = (n % 998244353 * fact[n] % 998244353) % 998244353;
  long long int sum = 0;
  for (long long int i = 1; i < n; i++)
    sum = (sum % 998244353 + (inverse(fact[i])) % 998244353) % 998244353;
  sum = (sum % 998244353 * (fact[n] % 998244353)) % 998244353;
  cout << (ans - sum + 998244353) % 998244353 << "\n";
}
