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.*;
import java.util.*;

public class Main {

	public static StringTokenizer ST;
	public static BufferedReader BR;
	public static PrintWriter PW;
	
	public static String next() {
		while(ST == null || !ST.hasMoreTokens()) {
			try {
				ST = new StringTokenizer(BR.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
				// TODO: handle exception
			}
		}
		return ST.nextToken();
	}
	
	public static int nextInt() {
		return Integer.parseInt(next());
	}
	
	public static long nextLong() {
		return Long.parseLong(next());
	}
	
	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
//		BR = new BufferedReader(new FileReader("1001.txt"));
		BR = new BufferedReader(new InputStreamReader(System.in));
		PW = new PrintWriter(new OutputStreamWriter(System.out));
		
		int n = nextInt();
		long M = 998244353;
		long[] fac = new long[n+1];
		long[] inv = new long[n+1];
		
		inv[1] = 1;
		for(int i = 2; i <= n; i++)
			inv[i] = M - M/i * inv[(int)M%i] % M;
		fac[0] = inv[0] = 1;
		for(int i = 1; i <= n; i++) {
			fac[i] = fac[i-1] * i % M;
			inv[i] = inv[i-1] * inv[i] % M;
		}
		
		long ans = 0;
		for(int l = 2; l <= n; l++) {
			ans = (ans +  (long)(n - l) * (l - 1) % M * inv[l]) % M;
		}
		ans = (ans + 1) * fac[n] % M;
		
		PW.println(ans);
		
		PW.close();
	}

}
