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). 
// package codeforce.Training1900;

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.Scanner;
//
public class DivideandSum {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
//        int t = sc.nextInt();
        int t = 1;
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static int mod = 998244353;
    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        int[] arr = new int[2 * n];
        for (int i = 0; i < 2 * n; i++) {
            arr[i] = in.nextInt();
        }
        ruffleSort(arr);
        long sum = 0;
        for (int i = 0; i < n; i++) {
            sum -= arr[i];
        }
        for (int i = n; i < 2 * n; i++) {
            sum += arr[i];
        }
        sum %= mod;
        Combinatorial comb = new Combinatorial(mod, 2 * n + 2);
        out.println(((sum * comb.ncr(2 * n, n)) % mod));
    }

    static class Combinatorial {

        int MOD;
        int n;
        long[] fact;
        long[] inv;
        long[] ifact;
        public Combinatorial(int mod, int n) {
            this.MOD = mod;
            this.n = n;
            fact = new long[n + 1];
            inv = new long[n + 1];
            ifact = new long[n + 1];
            inv[1] = 1;
            for (int i = 2; i < n; ++i) {
                inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
            }
            fact[0] = ifact[0] = 1;
            for (int i = 1; i < n; ++i) {
                fact[i] = fact[i - 1] * i % MOD;
                ifact[i] = ifact[i - 1] * inv[i] % MOD;
            }

        }

        long ncr(int n, int m){
            if (n < m || m < 0) return 0;
            return fact[n] * ifact[m] % MOD * ifact[n - m] % MOD;
        }
    }



    // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
    // worst case since it uses a version of quicksort. Although this would never
    // actually show up in the real world, in codeforces, people can hack, so
    // this is needed.
    static void ruffleSort(int[] a) {
        //ruffle
        int n=a.length;
        Random r=new Random();
        for (int i=0; i<a.length; i++) {
            int oi=r.nextInt(n), temp=a[i];
            a[i]=a[oi];
            a[oi]=temp;
        }

        //then sort
        Arrays.sort(a);
    }

}
