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;
const long long MOD = 998244353;
map<long long, long long> memo;
vector<long long> factorial;
long long power(long long a, long long x) {
  if (x == 0) return 1;
  long long mid = power(a, x / 2);
  long long res = (mid * mid) % MOD;
  if (x % 2) res = (res * a) % MOD;
  return res;
}
long long inverse(long long a) {
  if (!memo.count(a)) {
    memo[a] = power(a, MOD - 2);
  }
  return memo[a];
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  factorial.resize(n + 47);
  factorial[0] = 1;
  for (long long i = 1; i < n + 47; i++)
    factorial[i] = (factorial[i - 1] * i) % MOD;
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    long long curr = factorial[n];
    curr = (curr * (factorial[i] - 1)) % MOD;
    curr = (curr * inverse(factorial[i])) % MOD;
    res = (res + curr) % MOD;
  }
  cout << res << "\n";
}
