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.*;
import java.awt.image.BandedSampleModel;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.Scanner;
public class D {
    static long modInverse(long a, long m)
    {
        long g = gcd(a, m);

        {
            // If a and m are relatively prime, then modulo inverse
            // is a^(m-2) mode m
            //System.out.println("Modular multiplicative inverse is " +
                    //power(a, m - 2, m);
        }
        return power(a, m - 2, m);
    }

    // To compute x^y under modulo m
    static long power(long x, long y, long m)
    {
        if (y == 0)
            return 1;

        long p = power(x, y / 2, m) % m;
        p = (p * p) % m;

        if (y % 2 == 0)
            return p;
        else
            return (x * p) % m;
    }

    // Function to return gcd of a and b
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static void sort(long[] A){

        int n = A.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            long tmp = A[i];
            int randomPos = i + rnd.nextInt(n-i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int cases = 1;

        long mod=998244353l;
        for(int i=0;i<cases;i++)
        {
            int n = sc.nextInt();
             long arr[] = new long[2*n];
            for(int j=0;j<2*n;j++)
            {
                arr[j]=sc.nextLong();
            }

            sort(arr);
            long sum=0;
            for(int j=n;j<2*n;j++)
            {
                sum=sum+arr[j];
            }

            for(int j=0;j<n;j++)
            {
                sum=sum-arr[j];
            }
            sum=sum%mod;

            long ans=sum;
            for(int j=n+1;j<=2*n;j++)
            {
                ans=(ans*j)%mod;
            }

            for(int j=1;j<=n;j++)
            {
                ans=(ans*modInverse(j,mod))%mod;
            }

            System.out.println(ans%mod);
        }

    }
}
