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.io.*;
import java.util.*;
import java.lang.*;
import java.awt.*;
import java.awt.geom.*;
import java.math.*;
import java.text.*;
import java.math.BigInteger.*;
import java.util.Arrays; 

public class   Goodbye2018
{
  BufferedReader in;
  StringTokenizer as;
  long reversefact[]  = new long[1000001];
  long mod = 998244353 ;
  ArrayList < Integer > [] ar;
  
  public static void main (String[] args)
  {
    new Goodbye2018  ();
  }
  
  
  public int GCD(int a, int b) {
    if (b==0) return a;
    return GCD(b,a%b);
  }

  public long factorial(long number)
  {
    long total = 1;
    for(long x = 2;x<=number;x++)
    {
      total = (total * x) % mod;
    }
    return total;
  }
  
  public Goodbye2018  ()
  {
    try
    {
      
      in = new BufferedReader (new InputStreamReader (System.in));
      long a = nextInt();
      long total = factorial(a);
    //  System.out.println(total);
      long ans = ((total-1) * (a-2)) % mod;
      ans = (ans + total) % mod;
      long tt = 1;
      for(int x = (int)a;x>0;x--)
      {
        tt = (tt * x) % mod;
        reversefact[x] = tt;
      }
      //System.out.println(ans);
      //System.out.println(reversefact[3]);
      for(int x = 2;x<a;x++)
      {
        ans -= reversefact[x+1];
        ans += 1;
        ans = ans % mod;
        
      }
     // System.out.println(ans);
      System.out.println((ans + mod) % mod);
      
    }
    
    catch(IOException e)
    {
      
    }
  }
  
  
  
  
  
  
  
  String next () throws IOException
  {
    while (as == null || !as.hasMoreTokens ())
    {
      as = new StringTokenizer (in.readLine ().trim ());
    }
    
    
    return as.nextToken ();
  }
  
  
  
  long nextLong () throws IOException
  {
    return Long.parseLong (next ());
  }
  
  
  int nextInt () throws IOException
  {
    return Integer.parseInt (next ());
  }
  
  
  double nextDouble () throws IOException
  {
    return Double.parseDouble (next ());
  }
  
  
  String nextLine () throws IOException
  {
    return in.readLine ().trim ();
  }
}