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.*;

public class Main {
    
    
    static InputReader in;
    static PrintWriter out;
        
    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }
    
    static int N = (int)1e6;
    static class Task{
        
    	/*
    	 * i = 1~n-1
    	 * A(n,i)*( (n-i)! - 1);
    	 * n!/(n-i)!
    	 */
        int n;
        long fac[] = new long[N+10];
        long rfac[] = new long[N+10];
        long MOD = (int)998244353;
        
        long _pow(long x,long y) {
        	long ans = 1;
        	while (y>0) {
        		if (y%2==1) {
        			ans = (ans * x)%MOD;
        		}
        		x = x*x%MOD;
        		y = y/2;
        	}
        	return ans;
        }
        
        public void solve(InputReader in,PrintWriter out) {
        	n = in.nextInt();
        	fac[0] = 1;
        	for (int i = 1;i <= N;i++)
        		fac[i] = (1l*fac[i-1]*i)%MOD;
        	rfac[N] = _pow(fac[N],MOD-2);
        	for (int i = N-1;i>=0;i--)
        		rfac[i] = 1l*rfac[i+1]*(i+1)%MOD;
        	long ans = 0;
        	for (int i = 1;i <= n-1;i++) {
        		long temp = fac[n]*rfac[n-i]%MOD*(fac[n-i]-1)%MOD;
        		if (temp<0) temp+=MOD;
        		ans = (ans + temp)%MOD;
        	}
        	ans = (ans + fac[n])%MOD;
        	out.println(ans);
        }
    }

    

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;
        
        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }
        
        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}