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.
import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class D {
  public static int mod = 998244353;
  public static int[] fact;
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        // Scanner scan = new Scanner(System.in);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(bf.readLine());

        fact = new int[1000001];
        fact[0] = 1;
        for(int i=1; i<1000001; i++) {
          fact[i] = (int)(1L*fact[i-1]*i % mod);
        }


        long ans = 1;
        ans += 1L*n*fact[n] % mod;
        ans %= mod;
        for(int i=1; i<=n; i++) {
          ans -= 1L*fact[n-i]*choose(n, i) % mod;
          ans += mod;
          ans %= mod;
        }
        out.println(ans);



        out.close(); System.exit(0);
    }
    public static int exp(int base, int e) {
      if(e == 0) return 1;
      if(e == 1) return base;
      int val = exp(base, e/2);
      int ans = (int)(1L*val*val % mod);
      if(e % 2 == 1)
        ans = (int)(1L*ans*base % mod);
      return ans;
    }
    public static int inv(int base) {
      return exp(base, mod-2);
    }
    public static long frac(int p, int q) {
      return (1L*p*inv(q) % mod);
    }
    public static long choose(int a, int b) {
      if(a<b) return 0;
      if(b < 0) return 0;
      return frac(fact[a],(int)(1L*fact[b]*fact[a-b] % mod));
    }
}
