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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Solve8 {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(System.out);
        new Solve8().solve(pw);
        pw.flush();
        pw.close();
    }

    public void solve(PrintWriter pw) throws IOException {
        FastReader sc = new FastReader();
        int n = sc.nextInt();
        Integer[] a = new Integer[2 * n];
        for (int i = 0; i < 2 * n; i++) {
            a[i] = sc.nextInt();
        }
        Arrays.sort(a);
        final int MOD = 998244353;
        long s = 0;
        for (int i = 0; i < n; i++) {
            s = (s - a[i]) % MOD;
        }
        for (int i = n; i < 2 * n; i++) {
            s = (s + a[i]) % MOD;
        }
        long u = 1, d = 1;
        for (int i = 2; i <= 2 * n; i++) {
            u = (u * i) % MOD;
        }
        for (int i = 2; i <= n; i++) {
            d = (d * i) % MOD;
        }
        d = (d * d) % MOD;
        BigInteger modInv = BigInteger.valueOf(d);
        modInv = modInv.modInverse(BigInteger.valueOf(MOD));
        pw.println((s * ((u * modInv.longValue()) % MOD)) % MOD);
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

        public FastReader(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
            }
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                }
            }
            return st.nextToken();
        }

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

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

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

        public String nextLine() {
            try {
                return br.readLine();
            } catch (Exception e) {
            }
            return null;
        }

        public boolean hasNext() throws IOException {
            if (st != null && st.hasMoreTokens()) {
                return true;
            }
            String s = br.readLine();
            if (s == null || s.isEmpty()) {
                return false;
            }
            st = new StringTokenizer(s);
            return true;
        }
    }
}
