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.*;
import java.math.BigInteger;
import java.util.Map.Entry;

import static java.lang.Math.*;

public class D extends PrintWriter {

	long mod = 998244353;

	void run() {

		int n = nextInt();

		long[] f = new long[n + 1];
		f[0] = 1;
		for (int i = 1; i <= n; i++) {
			f[i] = (f[i - 1] * i) % mod;
		}

		long[] a = new long[n + 1];
		a[0] = 0;
		for (int i = 1; i <= n; i++) {
			a[i] = (a[i - 1] * i + 1) % mod;
		}

		long diff = (f[n] - a[n - 1] + mod) % mod;

		println((diff * n) % mod);

	}

	boolean skip() {
		while (hasNext()) {
			next();
		}
		return true;
	}

	int[][] nextMatrix(int n, int m) {
		int[][] matrix = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				matrix[i][j] = nextInt();
		return matrix;
	}

	String next() {
		while (!tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(nextLine());
		return tokenizer.nextToken();
	}

	boolean hasNext() {
		while (!tokenizer.hasMoreTokens()) {
			String line = nextLine();
			if (line == null) {
				return false;
			}
			tokenizer = new StringTokenizer(line);
		}
		return true;
	}

	int[] nextArray(int n) {
		int[] array = new int[n];
		for (int i = 0; i < n; i++) {
			array[i] = nextInt();
		}
		return array;
	}

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

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

	double nextDouble() {
		return Double.parseDouble(next());
	}

	String nextLine() {
		try {
			return reader.readLine();
		} catch (IOException err) {
			return null;
		}
	}

	public D(OutputStream outputStream) {
		super(outputStream);
	}

	static BufferedReader reader;
	static StringTokenizer tokenizer = new StringTokenizer("");
	static Random rnd = new Random();
	static boolean OJ;

	public static void main(String[] args) throws IOException {
		OJ = System.getProperty("ONLINE_JUDGE") != null;
		D solution = new D(System.out);
		if (OJ) {
			reader = new BufferedReader(new InputStreamReader(System.in));
			solution.run();
		} else {
			reader = new BufferedReader(new FileReader(new File(D.class.getName() + ".txt")));
			long timeout = System.currentTimeMillis();
			while (solution.hasNext()) {
				solution.run();
				solution.println();
				solution.println("----------------------------------");
			}
			solution.println("time: " + (System.currentTimeMillis() - timeout));
		}
		solution.close();
		reader.close();
	}
}