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.

// 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();
    }
}
