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.
//make sure to make new file!
import java.io.*;
import java.util.*;

public class DGB{

   public static long MOD = 998244353;
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int n = Integer.parseInt(f.readLine());
      
      long[] fac = new long[n+1];
      
      fac[0] = 1L;
      
      for(int k = 1; k <= n; k++){
         fac[k] = (fac[k-1] * k+MOD)%MOD;
      }
      
      long[] ifac = new long[n+1];
      
      ifac[1] = 1L * n;
      
      for(int k = 1; k < n; k++){
         ifac[k+1] = (ifac[k] * (n-k) + MOD)%MOD;
      }
      
      long answer = fac[n];
      
      for(int k = 1; k < n-1; k++){
         if(fac[n-k] == 0) out.println(k);
         long i = ifac[k];
         
         
         //answer = (answer + fac[n] - (fac[n]/fac[n-k]) + MOD)%MOD;
         answer = (answer + fac[n] - i + MOD)%MOD;
      
      }
      
      out.println((answer+MOD)%MOD);
      
      
      
      out.close();
   }
   
}