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>
const long long mod = 998244353;
using namespace std;
long long a[5000005], f[5000005];
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long ret = gcd(b, a % b, x, y);
    long long tmp = x;
    x = y;
    y = tmp - a / b * y;
    return ret;
  }
}
long long inv(long long a) {
  long long x, y;
  gcd(a, mod, x, y);
  return (x % mod + mod) % mod;
}
void solve() {
  f[0] = f[1] = 1;
  for (long long i = 2; i < 5000005; i++) {
    f[i] = i * f[i - 1] % mod;
  }
  long long n, m, k, x = 0, y = 0, c = 0, q, ans = 0;
  cin >> n;
  x += f[n];
  for (long long i = 1; i < n; ++i) {
    x += f[n] - f[n] * inv(f[i]);
    x %= mod;
  }
  x = (x + mod) % mod;
  cout << x << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
