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.
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class D_CF {

    public static void main(String[] args) {
        FastScanner58 fs = new FastScanner58();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        // for (int tc = 0; tc < t; tc++) {
        for (int tc = 0; tc < t; tc++) {
            long f = 1;
            long e = 1;
            long d = 0;
            int n = fs.ni();
            if (n==1) {
                pw.println(1);
                continue;
            }
            long[] ref = new long[n];
            for (int i = n-1; i > 0; i--) {
                f *= i;
                f %= (long) (998244353);
                ref[i-1] = f;
            }
            long res = ref[0]*n;
            for (int i = 1; i < n; i++) {
                e *= i;
                e %= (long) (998244353);
                long g = ((e - 1));
                d %= (long) (998244353);
                if (i!=n-1) g *= ref[i];
                d += g;
            }
            for (int i = 0; i < n; i++) {
                res += d;
                res %= (int) (998244353);
            }

            //pw.println(d);
            pw.println(res);
        }
        pw.close();
    }
}
//1 2 3
//1 3 2
// 2 1 3
// 2 3 1
// 3 1 2
// 3 2 1

//1 2 3 4
//1 2 4 3
// 1 3 2 4
//1 3 4 2
// 1 4 2 3
//1 4 3 2
class FastScanner58 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner58() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

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

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

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

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

class UnionFind17 {

    int[] id;

    public UnionFind17(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}
