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.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine());
        Integer a[]=new Integer[2*n];
        int i,N=300000; long f[]=new long[N+1],M=998244353;
        f[0]=1;
        for(i=1;i<=N;i++)
        f[i]=f[i-1]*i%M;

        long ans=nCr(2*n,n,f,M);
        String s[]=bu.readLine().split(" ");
        for(i=0;i<2*n;i++)
        a[i]=Integer.parseInt(s[i]);
        Arrays.sort(a);
        long sum=0;
        for(i=0;i<2*n;i++)
        if(i<n) sum-=a[i];
        else sum+=a[i];
        sum=(sum+M)%M;
        ans=ans*sum%M;
        System.out.print(ans);
    }

    static long nCr(int n,int r,long f[],long M)
    {
        return f[n]*power(f[r],M-2,M)%M*power(f[n-r],M-2,M)%M;
    }

    static long power(long a,long b,long M)
    {
        long res=1;
        while(b!=0)
        {
            if(b%2==1) res=res*a%M;
            b>>=1;
            a=a*a%M;
        }
        return res;
    }
}

