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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;
public class PCTest {
    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();
        long MOD = 998244353;
        long[][] prep = nckPrep(N, MOD);
        long out = 0L;
        for(int i = 1; i <= N-2; i++) {
            long temp = (nck(N, i, MOD, prep) * prep[i][0])%MOD;
            temp *= (prep[N-i][0]-1L+MOD);
            temp %= MOD;
            out= (out + temp) %MOD;
        }
        out = (out + prep[N][0])%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[][] nckPrep(int n, long MOD) {
        long[][] ret = new long[n+1][3];
        ret[0][0] = ret[1][0] = 1L;
        ret[0][2] = ret[1][2] = 1L;
        ret[0][1] = ret[1][1] = 1L;
        for(int i = 2; i <= n; i++) {
            ret[i][0] = (ret[i-1][0]*(long)i) % MOD;
            //This is another way to do modular inverse, distinct from those used above
            //it requires using previously computed modular inverses and is much faster than above
            long q = MOD/i;
            int r = (int)MOD%i;
            ret[i][1] = (-((q*ret[r][1])% MOD) + MOD) % MOD;
            ret[i][2] = (ret[i-1][2] * ret[i][1]) % MOD;
        }
        return ret;
    }

        static long nck(int n, int k, long MOD, long[][] prep) {
                 if (n < 0 || k > n) return 0L;
                 long ret = prep[n][0];
                 ret = (ret * prep[k][2]) % MOD;
                 ret = (ret * prep[n-k][2]) % MOD;
                 return ret;
            }
    
    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;
        }
    }
}
