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.math.BigInteger;
import java.text.DecimalFormat;
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int mod = 998244353;
        ArrayList<Long> fact = new ArrayList();
        fact.add(1l);
        int f = 1;
        for(int i=2;i<=n;i++){
            fact.add((long)i);
        }
        SegmentTree st = new SegmentTree(n);
        st.insert(fact);
        long sum = st.getMul(1,n);
        sum%=mod;
        for(int i=2;i<n;i++){
            //System.out.println( (st.getMul(i+1,n)-1)*(n-i));
            sum+= ((st.getMul(i+1,n)%mod)*(i-1)%mod)*(n-i)%mod;
            sum%=mod;
        }
        System.out.println(sum);
    }
}

class SegmentTree {
    ArrayList<Long> tree;
    int s;
    int mod = 998244353;
    public SegmentTree(int n) {
        for (s = 1; s < n; s *= 2);
        tree = new ArrayList<Long>(s * 2);
        tree.add(1l);
        for (int i = 1; i < s + s; i++)
            tree.add(1l);
    }

    void insert(ArrayList<Long> d) {
        for (int i = s; i < s + d.size(); i++)
            tree.set(i, d.get(i - s));
        for (int i = s - 1; i >= 1; i--)
            tree.set(i, tree.get(i * 2)*tree.get(i * 2 + 1)%mod);
    }

    long getMul(int Left, int Right) {
        int l = Left + s - 1, r = Right + s - 1;
        long rval = 1;
        while (l <= r) {
            if (l % 2 == 0) l /= 2;
            else {
                rval *= tree.get(l);
                rval %= mod;
                l = (l / 2) + 1;
            }
            if (r % 2 == 1) r /= 2;
            else {
                rval *=tree.get(r);
                rval %=mod;
                r = (r / 2) - 1;
            }
        }
        return rval;
    }
}
