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.
/**
 * ******* 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();
        }
    }

}
