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.util.*;
import java.io.*;
import java.awt.Point;
import java.math.BigInteger;

public class noway {
	
	static long MOD = 998244353;
	
	public static void main(String[] args) throws Exception {
		FastIO sc = new FastIO(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		long facts[] = new long[1000001];
		facts[0] = 1;
		for(int i = 1; i < facts.length; i++) {
			facts[i] = facts[i-1]*i;
			facts[i] %= MOD;
		}
		
		long[] calc = new long[1000001];
		calc[1] = 1;
		for(int i = 1; i < 1000000; i++) {
			calc[i+1] = (calc[i]+facts[i]-1)*(i+1);
			calc[i+1] %= MOD;
		}
		
		pw.println(calc[sc.nextInt()]);
		pw.close();
	}

	static class FastIO {

		//Is your Fast I/O being bad?

		InputStream dis;
		byte[] buffer = new byte[1 << 17];
		int pointer = 0;

		public FastIO(String fileName) throws Exception {
			dis = new FileInputStream(fileName);
		}

		public FastIO(InputStream is) throws Exception {
			dis = is;
		}

		int nextInt() throws Exception {
			int ret = 0;

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}

			return (negative) ? -ret : ret;
		}

		long nextLong() throws Exception {
			long ret = 0;

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}

			return (negative) ? -ret : ret;
		}

		byte nextByte() throws Exception {
			if (pointer == buffer.length) {
				dis.read(buffer, 0, buffer.length);
				pointer = 0;
			}
			return buffer[pointer++];
		}

		String next() throws Exception {
			StringBuffer ret = new StringBuffer();

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			while (b > ' ') {
				ret.appendCodePoint(b);
				b = nextByte();
			}

			return ret.toString();
		}

	}
}
