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.PrintWriter;
import java.util.StringTokenizer;

public class D {

    private static class FastReader {
        private static BufferedReader br;
        private static StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
    }

    public static int mod = 998244353;

    public static long add(long a, long b) {
        return (a % mod + b % mod) % mod;
    }

    public static long mult(long a, long b) {
        return ((a % mod) * (b % mod)) % mod;
    }

    public static long pow(long a, long b) {
        // fast exponentiation
        long c = 1;
        while (b > 0) {
            if (b % 2 == 1) {
                c = mult(c, a);
            }
            a = mult(a, a);
            b /= 2;
        }
        return c;
    }

    public static void main(String[] args) {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out);

        long n = in.nextLong();

        long m = n, a = n;
        for (long i = n - 2; i >= 1; i--) {
            a = add(a, mult(mult(m, i), n - i));
            m = add(m, mult(m, i));
        }

        out.println(a);
        out.close();
    }
}
