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;
const int N = (1e6) + 5;
long long fact[N];
long long GCD(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = GCD(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long ModInv(long long b) {
  long long x, y;
  GCD(b, MOD, &x, &y);
  return (x % MOD + MOD) % MOD;
}
long long ModDiv(long long a, long long b) {
  a %= MOD;
  long long inv = ModInv(b);
  return (inv * a) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, i, k, ans, temp;
  cin >> n;
  fact[1] = 1;
  for (i = 2; i <= n; i++) fact[i] = (fact[i - 1] * i) % MOD;
  ans = 1;
  for (k = 1; k <= n - 2; k++) {
    temp = k * n - k * k - k;
    temp %= MOD;
    temp *= fact[n];
    temp %= MOD;
    temp = ModDiv(temp, fact[n - k]);
    ans += temp;
    ans %= MOD;
  }
  ans += fact[n];
  ans %= MOD;
  ans--;
  ans += MOD;
  ans %= MOD;
  cout << ans << "\n";
  return 0;
}
