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.math.BigInteger;
import java.util.Scanner;

public class gb2018d {
	static final int MOD=998244353;
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		long[] fac=new long[n+1];
		fac[0]=1;
		for(int i=1;i<=n;i++) {
			fac[i]=(fac[i-1]*(long)i)%MOD;
		}
		long res=fac[n];
		
		for(int i=1;i<n;i++) {
//			System.out.println(res);
//			System.out.println(BigInteger.valueOf(fac[n-i]).modInverse(BigInteger.valueOf(MOD)));
			long inv=modInv(fac[n-i],MOD)%MOD;
//			System.out.println(inv);
			res=(res+(fac[n-i]-1)*(fac[n]*inv%MOD))%MOD;
		}
		System.out.println(res);
	}
	static int modInverse(long a, long m) 
    { 
		System.out.println(a);
        a = a % m; 
        for (int x = 1; x < m; x++) {
//        	System.out.println("...");
        	if ((a * x) % m == 1) return x;
        }
        return 1; 
    } 
	public static long choose2(long i) {
		return i*(i-1)/2;
	}
	static long modInv(long a, long b) {
		return 1 < a ? b - modInv(b % a, a) * b / a : 1;
	}

}