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 D {
    static long[] fact;
    static long A = 998244353l;
    static void gen(){
        fact = new long[1000002];
        fact[0] = 1;
        for(int i=1;i<fact.length;i++){
            fact[i] = fact[i-1]*i;
            fact[i]%=A;
        }
    }
     public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        gen();
        //f(n) = (n!-n) + f(n-1)*n
         long[] f = new long[1000001];
         for(int i=1;i<f.length;i++){
             f[i] = ((fact[i]-i)%A+A)%A + f[i-1]*i;
             f[i]%=A;
         }
         for(int i=0;i<f.length;i++){
             f[i]+=fact[i];
             f[i]%=A;
         }
         int n = sc.nextInt();
         pw.println(f[n]);
        pw.flush();
        pw.close();
    }
    static class FenwickTree { // one-based DS

        int n;
        int[] ft;

        FenwickTree(int size) { n = size; ft = new int[n+1]; }

        int rsq(int b) //O(log n)
        {
            int sum = 0;
            while(b > 0) { sum += ft[b]; b -= b & -b;}		//min?
            return sum;
        }

        int rsq(int a, int b) { return rsq(b) - rsq(a-1); }

        void point_update(int k, int val)	//O(log n), update = increment
        {
            while(k <= n) { ft[k] += val; k += k & -k; }		//min?
        }
        int get(int v){
            int sum =0;
            int id=0;
            for (int i =22;i>=0;i--){
                if (id+(1<<i)>n)
                    continue;
                if (sum+ft[id+(1<<i)]<v){
                    id+=(1<<i);
                    sum+=ft[id];
                }
            }
            return id+1;
        }
        public int findKthmin(int k){
            int a = 1;
            int b = n;
            int mid = 0;
            int res= 1;
            while (a<=b){
                mid = (a+b)/2;
                if(this.rsq(mid)>=k){
                    res =mid;
                    b = mid-1;
                }else{
                    a = mid+1;
                }
            }
            return res;
        }
    }
    static class Pair implements Comparable<Pair>{
        int x;
        int y;
        public Pair(int x,int y){
            this.x= x;
            this.y = y;
        }
        public int compareTo(Pair p){
            if(this.x==p.x)
                return Long.compare(this.y,p.y);
            return -Long.compare(this.x,p.x);
        }
        public String toString(){
            return x+ " " + y;
        }
    }
    static class Pair2 implements Comparable<Pair2>{
        int x;
        int y;
        int index;
        public Pair2(int x,int y){
            this.x= x;
            this.y = y;
            this.index = index;
        }
        public int compareTo(Pair2 p){
            if(this.y==p.y)
                return Long.compare(p.x,this.x);
            return Long.compare(this.y,p.y);
        }

    }
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();

        }
    }
}