Problem: 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 is 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. 

You are given n. Find the number of distinct subarrays of P. 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 distinct subarrays, modulo 998244353.

Examples

Input


2


Output


8


Input


10


Output


19210869

Note

In the first example, the sequence P = [1, 2, 2, 1]. It has eight distinct subarrays: [1], [2], [1, 2], [2, 1], [2, 2], [1, 2, 2], [2, 2, 1] and [1, 2, 2, 1]. 

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Wolfgang Beyer
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        int MOD = 998244353;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            if (n == 1) {
                out.println(1);
                return;
            }
            long result = Combinatorics.factorial(n, MOD);
            result *= n - 1;
            result %= MOD;

            long temp = 1;
            for (int i = n; i > 2; i--) {
                temp *= i;
                temp %= MOD;
                result -= temp;
                result = (result + MOD) % MOD;
            }
            out.println(result);
        }

    }

    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                    //tok = new StringTokenizer(in.readLine(), ", \t\n\r\f"); //adds commas as delimeter
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }

    static class Combinatorics {
        long[] fact;
        long[] invFact;
        int max;
        int mod;

        public Combinatorics(int modul, int m) {
            //modul should be prime

            mod = modul;
            max = m;
            fact = new long[max];
            invFact = new long[max];
            fact[1] = 1;
            fact[0] = 1;
            invFact[1] = 1;
            invFact[0] = 1;
            for (int i = 2; i < max; i++) {
                fact[i] = fact[i - 1] * i % mod;
            }

            invFact[max - 1] = modInverse(fact[max - 1], mod);
            for (int i = max - 2; i >= 0; i--) {
                invFact[i] = ((i + 1) * invFact[i + 1]) % mod; //.. 1 / i! = (i + 1) * 1 / (i + 1)!
            }
        }

        public Combinatorics(int modul) {
            // modul should be prime

            this(modul, 3000000);
        }

        public static long factorial(int n, int mod) {
            long result = 1;
            for (int i = 1; i <= n; i++) {
                result *= i;
                result = result % mod;
            }
            return (result);
        }

        public static long modPow(long a, long x, long p) {
            //calculates a^x mod p in logarithmic time.
            long res = 1;
            while (x > 0) {
                if (x % 2 != 0) {
                    res = (res * a) % p;
                }
                a = (a * a) % p;
                x /= 2;
            }
            return res;
        }

        public static long modInverse(long a, long p) {
            //calculates the modular multiplicative of a mod m.
            //(assuming p is prime).
            return modPow(a, p - 2, p);
        }

    }
}