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 {
	
	
	private void solve() throws Exception {
		long n = nextInt()-1;
		long T = n;
		long res = n+1L;
		for(long i = 2; i <= n; i++) {
			T = ( T * (n-i+3) ) %mod;
			T = ( T * (n-i+1) ) %mod;
			T = ( T * inverse(n-i+2) ) %mod;
			res = (res + (T*i)%mod)%mod;
		}
		out.println(res);
		
	}
	
	long mod = 998244353L;
	
	long inverse(long a) {
		long i = mod, v = 0, d = 1;
		while (a>0) {
			long t = i/a, x = a;
			a = i % x;
			i = x;
			x = d;
			d = v - t*x;
			v = x;
		}
		v %= mod;
		return (v+mod)%mod;
	}

	public static void main(String[] args) {
		new Thread(null, new Runnable() {
            public void run() {
                new Main().run();
            }
        }, "1", 1 << 27).start();
	}	

	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer tokenizer;

	public void run() {
		try {
			long start = System.nanoTime();
			in = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
			System.err.println((System.nanoTime()-start)/1E9);
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	private long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	private float nextFloat() throws IOException {
		return Float.parseFloat(nextToken());
	}

	private String nextLine() throws IOException {
		return new String(in.readLine());
	}

	private String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(in.readLine());
		}
		return tokenizer.nextToken();
	}


}