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.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);
        }

    }
}

