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 long mod = 998244353;
    public static void main(String[] args) throws IOException{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = pint(in);
        int[] a = new int[2 * n];
        long[] fact = new long[2 * n + 1];
        fact[0] = 1;
        for(int i = 1; i <= 2 * n; i++) {
        	fact[i] = (fact[i - 1] * i) % mod;
        }
        StringTokenizer st = new StringTokenizer(in.readLine());
        for(int i = 0; i < 2 * n; i++) {
        	a[i] = pint(st);
        }
        Arrays.sort(a);
        long s = 0;
        for(int i = 0; i < n; i++) {
        	s -= a[i]; s %= mod;
        }
        for(int i = n; i < 2 * n; i++) {
        	s += a[i]; s %= mod;
        }
        if(s < 0) {s += mod;}
        s *= fact[2 * n]; s %= mod;
        long ie = pow(fact[n], mod - 2);
        s *= ie; s %= mod;
        s *= ie; s %= mod;
        System.out.println(s);
    }
    static long pow(long a, long n) {
    	if(n == 0) {return 1;}
    	long ans = pow(a, n / 2); ans *= ans; ans %= mod;
    	if(n % 2 == 1) {ans *= a; ans %= mod;}
    	return ans;
    }
    static int pint(BufferedReader in) throws IOException {return Integer.parseInt(in.readLine());}
    static int pint(StringTokenizer st) {return Integer.parseInt(st.nextToken());}
}