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 Main {
	public static void main(String[] args)
	{
		//p_1098C();
		//p_445A();
		//p_431C();
		//p_706C();
		//p_44E();
		//p_106C();
		//p_1105A();
		//p_1105B();
		//p_1105C();
		//p_1105D();
		//p_1107A();
		//p_1107B();
		//p_1107C();
		//p_1091A();
		//p_1091B();
		//p_1091C();
		p_1091D();
	}
	private static void p_1091D()
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long s[]= new long[n];
		long sumS[] = new long[n];
		long n_to_1[] = new long [n];
		long result=0;
		if(n==1)
		{
			System.out.println(1);
			System.exit(0);
		}
		for(int i=0;i<n-1;i++)
		{
			if(i==0)
			{
				s[i]=n;
				sumS[i]=n;
				n_to_1[i] = n;
			}

			else
			{
				n_to_1[i] = (n_to_1[i-1] * (n-i))%998244353;
				s[i] = n_to_1[i];
				//for(int j=0;j<=i;j++)
				//	s[i] =(s[i]==0?(n-j):(s[i]*(n-j))%998244353);
				sumS[i] = s[i];
				//s[i] -=sumS[i-1];
				s[i] = (s[i]-sumS[i-1]>0?(s[i]-sumS[i-1]):s[i]-sumS[i-1]+998244353);
			}
		}
		for(int i=0;i<n-1;i++)
		{
			result += ((s[i] * (i + 1)) % 998244353);
			result = result%998244353;
		}
		System.out.println(result);
	}
}