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.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class D {
	static long MOD = 998244353;

	public static void main(String args[]) throws Exception {
		Scanner in = new Scanner(new BufferedInputStream(System.in));
		PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

		int N = in.nextInt();

		long fact = N;
		long ans = N * factorial(N) - N + 1;
		ans %= MOD;

		long sum = 0;
		for (int i = 1; i < N; i++) {
			long curr = fact - 1 - sum;
			curr = (curr + MOD) % MOD;

			ans -= (N - i) * curr;
			ans += MOD * N;
			ans %= MOD;

			fact *= (N - i);
			fact %= MOD;

			sum += curr;
			sum %= MOD;
		}
		out.println(ans);

		in.close();
		out.close();
	}

	static long factorial(int n) {
		if (n == 1) return 1;
		return n * factorial(n - 1) % MOD;
	}

}
