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.
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskD {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    long M = 998_244_353L;

    long n = Integer.parseInt(br.readLine());
    if (n==1) {
      System.out.println("1");
      return;
    }
    long[] facts = new long[(int)n];
    long[] invFacts = new long[(int)n];
    facts[0] = 1;
    for (int i=1; i<n; i++) {
      facts[i] = facts[i-1]*i;
      if (facts[i]>=M) {
        facts[i] %= M;
      }
      invFacts[i] = modInv(facts[i], M);
    }
    long sum = (n-1)*facts[(int)n-1];
    if (sum>=M)
      sum %= M;
    for (int i=2; i<=n-1; i++) {
      long sub = facts[(int)n-1]*invFacts[i];
      if (sub>=M){
        sub %= M;
      }
      sum -= sub;
      if (sum<0) {
        sum += M;
      }
    }
    sum *= n;
    if (sum>=M)
      sum %= M;
    System.out.println(sum);
    
  }
  static long modInv(long x, long m) { 
    return (egcd(x, m)[1] + m) % m; 
  }
  static long[] egcd(long a, long b) {
    if (b == 0) 
      return new long[] { a, 1, 0 };
    else {
      long[] ret = egcd(b, a % b);
      long tmp = ret[1] - ret[2] * (a / b);
      ret[1] = ret[2]; 
      ret[2] = tmp; 
      return ret; 
    } 
  }
  static long gcd(long a, long b) {
    return b == 0 ? (a<0?-a:a) : gcd(b, a % b); 
  }
}
