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 Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;
    int[] id, size;
    public static void main(String[] args){
        Main main = new Main();
        int t = 1;
        while(t-->0){
            main.solve();
        }
        out.flush();
    }
    void solve(){
        int n = sc.nextInt();
        long[] a = new long[2*n];
        for(int i=0; i<2*n; i++){
            a[i] = sc.nextLong();
        }
        Arrays.sort(a);
        int maxn = 2*n;
        long[] frac = new long[maxn+1];
        frac[0] = 1;
        for(int i=1; i<=maxn; i++){
            frac[i] = frac[i-1]*i%mod;
        }
        long comb = frac[maxn]*power(frac[n], mod-2)%mod*power(frac[n], mod-2)%mod; // C(2n, n)
        long ans = 0;
        for(int i=0; i<n; i++){
            // smaller part
            long temp = a[i]*comb%mod;
            ans = (ans+mod-temp)%mod;
        }
        for(int i=n; i<2*n;i++){
            // bigger part
            long temp = a[i]*comb%mod;
            ans = (ans+temp)%mod;
        }
        out.println(ans);

    }

    long power(long base, long p){
        long ans = 1;
        while(p>0){
            if(p%2==1) ans = ans*base%mod;
            base = base*base%mod;
            p /= 2;
        }
        return ans;
    }

}
