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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int mod = 998244353;
		int[] input = new int[2*n];
		for(int i = 0;i < input.length;i++) input[i] = s.nextInt();
		Arrays.sort(input);
		long first = 0;
		for(int i = 0;i < input.length/2;i++) first = (first%mod + input[i]%mod)%mod;
		long second = 0;
		for(int i = input.length/2;i < input.length;i++) second = (second%mod + input[i]%mod)%mod;
		second = (second - first)%mod;

		long n_2 = fac(2 * n)%mod;	
		long n_1 = fac(n)%mod;
		n_1 = (n_1 * n_1)%mod;

		long ans = (n_2) * (binExpo(n_1 , mod - 2 , mod));
		ans %= mod;
		ans = (ans * second)%mod;
		ans = (ans + mod) % mod;
		System.out.println(ans);
	}

	public static long fac(int n) {
		long ans = 1;
		int mod = 998244353;
		for(int i = 2;i <= n;i++) ans = (ans * i)%mod;

		return ans;
	}
	public static long binExpo(long d , int m , int mod) {

		long ans = 1;
		while(m != 0) {

			if(m % 2 == 1) {
				ans = (ans * d)%mod; // ans = 2
			}
			//			System.out.println(ans + " l");
			m /= 2;
			d = (d%mod * d%mod)%mod;
			//System.out.println(ans);

		}
		return ans;
	}

}
