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>
inline bool VISED(int &x) {
  bool t(x);
  x = 1;
  return t;
}
using namespace std;
const long long llinf = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn = int(1e6 + 10);
const int mod(998244353);
int n;
long long jie[maxn] = {1};
long long solve(int n) {
  long long ans(jie[n] * n % mod);
  ans = ans;
  long long p(n), res(n);
  for (long long len1 = n - 1; len1 >= 2; --len1) {
    res = res * len1 % mod;
    p = (p + res) % mod;
  }
  ans = ans - p;
  ans %= mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  int T(1), times(0);
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (++times, T--) {
    cin >> n;
    if (n == 1) {
      cout << 1 << endl;
      continue;
    }
    for (int i = 1; i <= n + 2; i++) jie[i] = jie[i - 1] * i % mod;
    cout << solve(n) << endl;
  }
  return 0;
}
