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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;
public class D {
    static final long MOD = 998244353;
    static long fact[];
    static long invfact[];
    static long inv[];
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        //dont forget overflow/mod;
        int N = scanner.nextInt();
        fact = new long[N+1];
        invfact = new long[N+1];
        inv = new long[N+1];
        fact[0] = 1L;
        fact[1] = 1L;
        invfact[0] = 1L;
        invfact[1] = 1L;
        inv[1] = 1L;
        for(int i = 2; i <= N; i++) {
            fact[i] = (fact[i-1]*(long)i)%MOD;
            //inv[i] = modInverse(i);
            long q = MOD/i;
            int r = (int)MOD%i;
            inv[i] = (-(q*inv[r])%MOD+MOD)%MOD;
            invfact[i] = (invfact[i-1]*inv[i])%MOD;
        }
        long out = 0L;
        for(int i = 1; i <= N-2; i++) {
            long temp = (nCr(N, i) * fact[i])%MOD;
            temp*= (fact[N-i]-1L+MOD);
            temp%=MOD;
            out= (out + temp) %MOD;
        }
        out = (out + fact[N])%MOD;
        System.out.println(out);
    }
    public static long modInverse(long x)
    {
        return modPow(x, MOD-2);
    }
    public static long modPow(long x, long y) {
        if (y == 0) return 1;
        long p = modPow(x, y / 2) %MOD;
        p = (p * p) % MOD;
        if (y % 2 == 0) return p;
        else return (x * p) % MOD;
    }
    static long nCr(int n,int r)
    {
        long res;
        if(n<0 || r>n)
            return 0;
        res = fact[n];
        res = (res*invfact[r])%MOD;
        res = (res*invfact[n-r])%MOD;
        return res;
    }
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}