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]. 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;

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

    static class DNewYearAndThePermutationConcatenation {
        long mod = 998244353;
        long[] fac;

        public void solve(int testNumber, ScanReader in, PrintWriter out) {
            int n = in.scanInt();
            fac = new long[(int) n + 5];
            fac[0] = 1;
            fac[1] = 1;
            for (int i = 2; i <= n; i++) {
                fac[i] = (fac[i - 1] * i) % mod;
            }


            long ans = fac[n];
            for (int i = n - 1; i >= 1; i--) {
                long temp = ((fac[i] - 1) * ncr(n, n - i)) % mod;
                temp = (temp * fac[n - i]) % mod;
                ans = (ans + temp) % mod;
            }

            out.println(ans);
        }

        long ncr(int n, int r) {
            long up = fac[(int) n];
            long dd = (fac[n - r] * fac[r]) % mod;
            return (up * CodeX.modinv(dd, mod)) % mod;
        }

    }

    static class CodeX {
        public static long power(long x, long y, long p) {
            long res = 1;
            x = x % p;
            while (y > 0) {
                if ((y & 1) != 0)
                    res = (res * x) % p;
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

        public static long modinv(long k, long mod) {
            return power(k, mod - 2, mod);
        }

    }

    static class ScanReader {
        private byte[] buf = new byte[4 * 1024];
        private int INDEX;
        private BufferedInputStream in;
        private int TOTAL;

        public ScanReader(InputStream inputStream) {
            in = new BufferedInputStream(inputStream);
        }

        private int scan() {
            if (INDEX >= TOTAL) {
                INDEX = 0;
                try {
                    TOTAL = in.read(buf);
                } catch (Exception e) {
                    e.printStackTrace();
                }
                if (TOTAL <= 0) return -1;
            }
            return buf[INDEX++];
        }

        public int scanInt() {
            int I = 0;
            int n = scan();
            while (isWhiteSpace(n)) n = scan();
            int neg = 1;
            if (n == '-') {
                neg = -1;
                n = scan();
            }
            while (!isWhiteSpace(n)) {
                if (n >= '0' && n <= '9') {
                    I *= 10;
                    I += n - '0';
                    n = scan();
                }
            }
            return neg * I;
        }

        private boolean isWhiteSpace(int n) {
            if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
            else return false;
        }

    }
}

