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.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
 
public class Main 
{
    static long power(long x, int y, int p)
    {
 
        // Initialize result
        long res = 1L;
 
        // Update x if it is more than or
        // equal to p
        x = x % p;
 
        while (y > 0) {
 
            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;
 
            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
 
        return res;
    }
 
    // Returns n^(-1) mod p
    static long modInverse(long n, int p)
    {
        return power(n, p - 2, p);
    }
 
    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r,
                             int p)
    {
 
          if (n<r) 
              return 0;
      // Base case
        if (r == 0)
            return 1;
 
        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        long[] fac = new long[n + 1];
        fac[0] = 1;
 
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;
 
        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
            % p;
    }
 
    public static void main(String[] args)
    {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int[] a=new int[2*n];
        for(int i=0;i<2*n;i++)
            a[i]=in.nextInt();
        Arrays.sort(a);
        int[] p=new int[n];
        int[] q=new int[n];
        int c=0;
        for(int i=0;i<2*n;i+=2)
        {
            p[c]=a[i];
            q[c]=a[i+1];
            c++;
        }
        long sum=0L;
        for(int i=0;i<n;i++)
        {
            sum+=Math.abs(p[i]-q[n-1-i]);
            sum%=998244353;
        }  
        long comb=nCrModPFermat(2*n,n,998244353);
        System.out.println((comb*sum)%998244353);
         
    }
} 


