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]. 
/**
 * ******* Created  on 25/4/20 12:31 PM*******
 */

import java.io.*;
import java.util.*;

public class D1091 implements Runnable {

    private static final int MAX = (int) (1E5 + 5);
    private static final int MOD = 998244353 ;
    private static final long Inf = (long) (1E14 + 10);
    private static final double eps = (double) (1E-9);

    private void solve() throws IOException {
        int n = reader.nextInt();


        long[] fact = new long[n+1];
        long[] invFact = new long[n+1];
        fact[0] = invFact[0]=1;

        for(int i=1;i<=n;i++){
            fact[i] = (fact[i-1]*i )%MOD;
            invFact[i] = pow(fact[i], MOD-2);
        }
        long ans = (n*fact[n])%MOD ;
        for(int i=1;i<n;i++){
            long val = (fact[n]*invFact[i] )%MOD;
            ans = (ans - val +MOD)%MOD;
        }
        writer.println(ans);
    }

    private long pow(long a, int b) {
        long ans =1;
        while (b >0){
            if(b%2 ==1)ans =(ans*a) %MOD;
            b>>=1;
            a= (a*a)%MOD;
        }
        return ans;
    }

    public static void main(String[] args) throws IOException {
        try (Input reader = new StandardInput(); PrintWriter writer = new PrintWriter(System.out)) {
            new D1091().run();
        }
    }

    StandardInput reader;
    PrintWriter writer;

    @Override
    public void run() {
        try {
            reader = new StandardInput();
            writer = new PrintWriter(System.out);
            solve();
            reader.close();
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }


    interface Input extends Closeable {
        String next() throws IOException;

        String nextLine() throws IOException;

        default int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        default long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        default double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        default int[] readIntArray() throws IOException {
            return readIntArray(nextInt());
        }

        default int[] readIntArray(int size) throws IOException {
            int[] array = new int[size];
            for (int i = 0; i < array.length; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        default long[] readLongArray(int size) throws IOException {
            long[] array = new long[size];
            for (int i = 0; i < array.length; i++) {
                array[i] = nextLong();
            }
            return array;
        }
    }

    private static class StandardInput implements Input {
        private final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        private StringTokenizer stringTokenizer;

        @Override
        public void close() throws IOException {
            reader.close();
        }

        @Override
        public String next() throws IOException {
            if (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
                stringTokenizer = new StringTokenizer(reader.readLine());
            }
            return stringTokenizer.nextToken();
        }

        @Override
        public String nextLine() throws IOException {
            return reader.readLine();
        }
    }

}
