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.lang.*;
import java.io.*;

//run without debug: ctrl+a
//clear terminal: ctrl+space

public class B_Divide_and_Sum
{   

     
     public static void main(String[] args) throws java.lang.Exception 
    {

          Scanner s = new Scanner(System.in);
          int n=s.nextInt();
          int []a=new int[2*n];for(int i=0;i<2*n;i++)a[i]=s.nextInt();
          long M=998244353;
          long ans=0;

         Arrays.sort(a);
          for(int i=0;i<n;i++)
          {
                   ans+=a[n+i]-a[i];
                   ans%=M;               
          }

          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]%= M;
        }
 
        //we need to find the modular inverse of fac[n]
        //the extended euclidean algorithm
 
        long invFacn = modInv(M,fac[n]);
        long result=fac[2*n]%M*invFacn%M*invFacn%M;
        ans=(ans*result)%M;

        System.out.println(ans);

      }
      
      static 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;

      }
}