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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    static long mod = 998244353;
    static int n;
    static long[] fac,inv;
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        init();
        long ans = fac[n];
        for(int i=1;i<=n-1;i++){
            long cur = fac[i]*comb(n,i)%mod*((fac[n-i]-1+mod)%mod)%mod;
            ans = (ans+cur)%mod;
        }
        System.out.println(ans);
    }
    static long comb(int big, int small){
        // choose m out of n. m<=n;
        if(small>big) return 0;
        long ans = fac[big]*inv[small]%mod*inv[big-small]%mod;
        return ans;
    }
    static void init(){
        fac = new long[n+1];
        inv = new long[n+1];
        fac[0] = 1;
        for(int i=1;i<=n;i++) fac[i] = fac[i-1]*i%mod;
        inv[0] = 1;
        inv[n] = power(fac[n],mod-2);
        for(int i=n-1;i>=1;i--) inv[i] = (i+1)*inv[i+1]%mod;
    }
    static long power(long base, long q){
        long ans = 1;
        while(q>0){
            if(q%2==1) ans = ans*base%mod;
            base = base*base%mod;
            q /= 2;
        }
        return ans;
    }
}
