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). 
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1445D extends PrintWriter {
	CF1445D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1445D o = new CF1445D(); o.main(); o.flush();
	}

	static final int MD = 998244353;
	long inv(int a) {
		return a == 1 ? 1 : inv(a - MD % a) * (MD / a + 1) % MD;
	}
	void main() {
		int n = sc.nextInt();
		int[] aa = new int[n * 2];
		for (int i = 0; i < n * 2; i++)
			aa[i] = sc.nextInt();
		aa = Arrays.stream(aa).boxed().sorted().mapToInt($->$).toArray();
		int a = 0;
		for (int i = 0; i < n; i++)
			a = (a + aa[n + i] - aa[i]) % MD;
		long f = 1, g = 1;
		for (int i = 1; i <= n * 2; i++) {
			f = f * i % MD;
			if (i == n)
				g = inv((int) f);
		}
		println(a * f % MD * g % MD * g % MD);
	}
}
