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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class D{
    static final long MOD = 998244353;
    static long[] fact = new long[1000000+7];
    static long[] invfact = new long[1000000+7];
    private static long inverse(long a){
        long b = MOD-2;
        long res = 1;
        while(b!=0){
            if((b&1)==1){
                res = (res*a)%MOD;
            }
            a = (a*a)%MOD;
            b >>= 1;
        }
        return res%MOD;
    }
    private static void factorial(){
        fact[1] = 1;
        invfact[1] = 1;
        for(int i=2; i<fact.length; i++){
            fact[i] = fact[i-1]*i;
            fact[i] %= MOD;
            invfact[i] = inverse(fact[i])%MOD;
        }
    }
    public static void main(String[] args){
        FS sc = new FS();
        PrintWriter pw = new PrintWriter(System.out);
        int n = sc.nextInt();
        factorial();
        long ans = 1;
        ans = (ans*n*fact[n])%MOD;
        for(int i=1; i<n; i++){
            long temp = (fact[n]*invfact[i])%MOD;
            ans = (ans-temp)%MOD;
        }
        pw.println(ans<0 ? ans + MOD : ans);
        pw.flush();
        pw.close();
    }

    static class FS{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
    
        String next(){
            while(!st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored){
                }
            }
            return st.nextToken();
        }
        
        int[] nextArray(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++){
                a[i] = nextInt();
            }
            return a;
        }
        
        long[] nextLongArray(int n){
            long[] a = new long[n];
            for(int i = 0; i < n; i++){
                a[i] = nextLong();
            }
            return a;
        }
        
        int nextInt(){
            return Integer.parseInt(next());
        }
        
        long nextLong(){
            return Long.parseLong(next());
        }
        
        double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}