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.util.*;


public class Main {


    public Main(int n, int[] numbers) {

        //we assume that the costs of all partitions are equal
        //Therefore, we only have to find the cost of one partition
        //and the number of possible partitions

        int[] part1 = new int[n];
        int[] part2 = new int[n];
        for (int i = 0; i < numbers.length/2; i++) {
            part1[i]=numbers[i];
            part2[i]=numbers[i+n];
        }

        Arrays.sort(part1);
        Arrays.sort(part2);


        long cost = 0;
        for (int i = 0; i < n; i++) {
            cost+=Math.abs(part2[n-1-i]-part1[i]);
        }
        long mod = 998244353;
        cost%= mod;


        //Calculate the number of partitions
        //we choose n out of 2n for the first partition
        //2n!/n!/n!

        long[] fac = new long[2*n+1];
        fac[0]=1;
        for (int i = 1; i < 2*n+1; i++) {
            fac[i]=fac[i-1]*i;
            fac[i]%= mod;
        }

        //we need to find the modular inverse of fac[n]
        //the the extended euclidean algorithm

        long invFacn = modInv(mod,fac[n]);
        long result = cost*fac[2*n]%mod*invFacn%mod*invFacn%mod;
        System.out.println(result);
    }



    public long modInv(long mod, long n) {
        long q = mod;
        long r = n;
        long c1 = 1;
        long d1 = 0;
        long c2 = 0;
        long d2 = 1;
        long x1;
        long x2;
        while (r!=0) {
             long tmpq = q/r;
             long tmpr = q%r;
            x1 = c1-(tmpq*d1);
            x2 = c2-(tmpq*d2);
            c1 = d1;
            c2 = d2;
            d1 = x1;
            d2 = x2;
            q = r;
            r = tmpr;
        }
        if (c2<0)
            c2+=mod;
        return c2;
    }





    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[2 * n];
        for (int i = 0; i < 2 * n; i++)
            a[i] = sc.nextInt();

       new Main(n, a);

      

    }

}
