You are given an array a of length 2n. Consider a partition of array a into two subsequences p and q of length n each (each element of array a should be in exactly one subsequence: either in p or in q).

Let's sort p in non-decreasing order, and q in non-increasing order, we can denote the sorted versions by x and y, respectively. Then the cost of a partition is defined as f(p, q) = ∑_{i = 1}^n |x_i - y_i|.

Find the sum of f(p, q) over all correct partitions of array a. Since the answer might be too big, print its remainder modulo 998244353.

Input

The first line contains a single integer n (1 ≤ n ≤ 150 000).

The second line contains 2n integers a_1, a_2, …, a_{2n} (1 ≤ a_i ≤ 10^9) — elements of array a.

Output

Print one integer — the answer to the problem, modulo 998244353.

Examples

Input


1
1 4


Output


6

Input


2
2 1 2 1


Output


12

Input


3
2 2 2 2 2 2


Output


0

Input


5
13 8 35 94 9284 34 54 69 123 846


Output


2588544

Note

Two partitions of an array are considered different if the sets of indices of elements included in the subsequence p are different.

In the first example, there are two correct partitions of the array a:

  1. p = [1], q = [4], then x = [1], y = [4], f(p, q) = |1 - 4| = 3; 
  2. p = [4], q = [1], then x = [4], y = [1], f(p, q) = |4 - 1| = 3. 



In the second example, there are six valid partitions of the array a: 

  1. p = [2, 1], q = [2, 1] (elements with indices 1 and 2 in the original array are selected in the subsequence p); 
  2. p = [2, 2], q = [1, 1]; 
  3. p = [2, 1], q = [1, 2] (elements with indices 1 and 4 are selected in the subsequence p); 
  4. p = [1, 2], q = [2, 1]; 
  5. p = [1, 1], q = [2, 2]; 
  6. p = [2, 1], q = [2, 1] (elements with indices 3 and 4 are selected in the subsequence p). 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class SolutionB extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                                            InputStreamReader(System.in));
        }

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

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

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

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

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

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        new Thread(null, new SolutionB(), "Main", 1 << 26).start();
    }

    public void run() {
        int t = 1;
        for (int i = 0; i < t; i++) {
            solve();
        }
        out.close();
    }

    static final long MODULO = 998244353;

    static private class GcdTriple {
        long d;
        long s;
        long t;

        public GcdTriple(long d, long s, long t) {
            this.d = d;
            this.s = s;
            this.t = t;
        }
    }

    public static GcdTriple extendedGcd(long a, long b) {
        if (b == 0) {
            return new GcdTriple(a, 1, 0);
        }
        GcdTriple result = extendedGcd(b, a % b);

        long tmp = result.s;
        result.s = result.t;
        result.t = tmp - (a / b) * result.t;
        return result;
    }

    //calculate b^-1 , so that (b * b^-1) == 1 under mod m;
    public static long modInverse(long b, long m) {
        GcdTriple inverseTriple = extendedGcd(b, m);

        if (inverseTriple.d != 1) {
            return -1;
        }
        return (inverseTriple.s % m + m) % m;
    }

    private static void solve() {
        int n = scanner.nextInt();
        Integer[] a = new Integer[2 * n];
        for (int i = 0; i < 2 * n; i++) {
            a[i] = scanner.nextInt();
        }
        Arrays.sort(a);

        long amountPartitions = 1;
        for (int i = n+1; i <= 2 * n; i++) {
            amountPartitions = (amountPartitions * i) % MODULO;
        }
        for (int i = 2; i <= n; i++) {
            amountPartitions = (amountPartitions * modInverse(i, MODULO)) % MODULO;
        }
        
        long sum = 0;
        for (int i = 0; i < n; i++) {
            sum = (sum + (a[i + n] - a[i]) * amountPartitions) % MODULO;
        }
        System.out.println(sum);
    }

    //REMINDERS:
    //- CHECK FOR INTEGER-OVERFLOW BEFORE SUBMITTING

    //- CAN U BRUTEFORCE OVER SOMETHING, TO MAKE IT EASIER TO CALCULATE THE SOLUTION
}