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.io.*;
public class q4{
    public static void main(String args[]) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        int arr[] = new int[2*n];
        for(int i=0;i<2*n;i++){
            arr[i]= Integer.parseInt(st.nextToken());
        }
        Arrays.sort(arr);
        long first = 0;
        long second = 0;
        for(int i=0;i<n;i++){
            first += arr[i];
        }
        for(int i=n;i<2*n;i++){
            second += arr[i];
        }
        long mod = 998244353;
        long ans = ((second-first +mod )%mod * bin(2*n,n,mod))%mod;
        System.out.println(ans);
        
    }
    public static long bin(long n,long k,long mod){
        if(k>n-k){
            k = n-k;
        }
        long res = 1;
        for(long i=0;i<k;i++){
            res = (res*(n-i))%mod;
            res = res*(power(i+1,mod-2,mod))%mod;
            
        }
        return res;
    }
    public static long power(long a,long b,long m){
        long res = 1;
        a = a%m;
        while(b>0){
            if(b%2==1){
                res = res*a %m;
            }
            a = a*a%m;
            b = b/2;
        }
        return res;
    }
}