Problem: 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 is 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. 

You are given n. Find the number of distinct subarrays of P. 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 distinct subarrays, modulo 998244353.

Examples

Input


2


Output


8


Input


10


Output


19210869

Note

In the first example, the sequence P = [1, 2, 2, 1]. It has eight distinct subarrays: [1], [2], [1, 2], [2, 1], [2, 2], [1, 2, 2], [2, 2, 1] and [1, 2, 2, 1]. 

Solution (in JAVA):

Solution:

// LM10: The next Ballon d'or
import java.util.*;
import java.io.*;
import java.math.*;
import javafx.util.Pair;
public class Main
{
    static class FastReader
    {
        private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public FastReader() { this(System.in); }public FastReader(InputStream is) { mIs = is;}
        public int read() {if (numChars == -1) throw new InputMismatchException();if (curChar >= numChars) {curChar = 0;try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}if (numChars <= 0) return -1; }return buf[curChar++];}
        public String nextLine(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isEndOfLine(c));return res.toString() ;}
        public String next(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isSpaceChar(c));return res.toString();}
        public long l(){int c = read();while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }long res = 0; do{ if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read();}while(!isSpaceChar(c));return res * sgn;}
        public int i(){int c = read() ;while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }int res = 0;do{if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read() ;}while(!isSpaceChar(c));return res * sgn;}
        public double d() throws IOException {return Double.parseDouble(next()) ;}
        public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; }
        public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; }
        public void scanIntArr(int [] arr){ for(int li=0;li<arr.length;++li){ arr[li]=i();}}
        public void scanLongArr(long [] arr){for (int i=0;i<arr.length;++i){arr[i]=l();}}
        public void shuffle(int [] arr){ for(int i=arr.length;i>0;--i) { int r=(int)(Math.random()*i); int temp=arr[i-1]; arr[i-1]=arr[r]; arr[r]=temp; } }
    }
    public static void main(String[] args)throws IOException {
        /*
inputCopy
3
outputCopy
9
inputCopy
4
outputCopy
56
inputCopy
10
outputCopy
30052700
         */
        PrintWriter pw = new PrintWriter(System.out);
        FastReader fr = new FastReader();
        long mod=998244353L;
        int n=fr.i();
        long [] dp=new long[n+1];
        dp[1]=1;
        if(n>1)
            dp[2]=2;
        if(n>2)
            dp[3]=9;
        long [] fact=new long [n+1];
        fact[0]=fact[1]=1;
        if(n>1)
            fact[2]=2;
        if(n>2)
            fact[3]=6;
        for(int ni=4;ni<=n;++ni)
        {
            fact[ni]=(ni*fact[ni-1])%mod;
            dp[ni]=(fact[ni] + (ni*(dp[ni-1]-1)))%mod;
        }
        pw.println(dp[n]);
        pw.flush();
        pw.close();
    }
}