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.lang.reflect.Array;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {
    
    private static final Scanner sc = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        //StringTokenizer st = new StringTokenizer(br.readLine());
        int mod = 998244353;
        int n = sc.nextInt();
        if (n==1){
            System.out.println(1);
            return;
        }
        long[] arr = new long[n];
        arr[1] = 1;
        for (int i=2; i<n; ++i){
            arr[i] = (arr[i-1]*(i+1))%mod;
            long z = i;
            z *= z;
            arr[i] += z;
            arr[i]%=mod;
        }
        long ans = 1;
        for (int i=2; i<=n; ++i){
            ans*=i;
            ans%=mod;
        }
        ans*=n;
        ans%=mod;
        ans-=arr[n-1];
        ans-=n-1;
        ans+=2*mod;
        ans%=mod;
        System.out.println(ans);
        
        sc.close();
        br.close();
        bw.close();
    }
    
}