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.util.*;

public class d1091 {
    static final long mod = 998244353;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        solve(n);
        // solve(3); // 9
        // solve(4); // 56
        // solve(10); // 30052700
        // solve(100); // 204570371
        // solve(200); // 796904570
        // solve(300); // 620337026
        // solve(10000); // 175554733
    }

    static void solve(int n) {

        // factorials
        long[] fac = new long[n + 1];
        fac[0] = fac[1] = 1;
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % mod;

        // mod inverse
        long[] inv = new long[n + 1];
        for (int i = 0; i <= n; i++) {
            inv[i] = modInverse(fac[i], mod);
        }

        // long ret = n * f[n] % mod;
        long ret = fac[n];

        for (int i = 1; i < n; i++) {
            ret = (ret + fac[n]) % mod;
            ret = (ret - (fac[n] * inv[i] % mod) + mod) % mod;
        }

        System.out.println(ret);
    }

    static long modInverse(long a, long m) {
        long m0 = m;
        long y = 0, x = 1;

        if (m == 1)
            return 0;

        while (a > 1) {
            // q is quotient
            long q = a / m;

            long t = m;

            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;

            // Update x and y
            y = x - q * y;
            x = t;
        }

        // Make x positive
        if (x < 0)
            x += m0;

        return x;
    }
}
