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.io.*; 
import java.util.*; 
import java.math.*; 

public class Main {
	static long mod=998244353;
	static long[] facts=new long[3_000_05];
	static long mul(long a, long b) {
		return a*b%mod;
	}
	
	static long add(long a, long b) {
		return (a+b)%mod;
	}
	
	static void precomp() {
		facts[0]=1;
		for (int i=1; i<facts.length; i++) facts[i]=mul(facts[i-1], i);
	}
	
	static long exp(long base, long pow) {
		if (pow==0) return 1;
		long half=exp(base, pow/2);
		if (pow%2==0) return mul(half, half);
		return mul(half, mul(half, base));
	}
	
	static long modInv(long x) {
		return exp(x, mod-2);
	}
	
	static long nCk(int n, int k) {
		if(n<k) return 0;
		return mul(facts[n], mul(modInv(facts[k]), modInv(facts[n-k])));
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
//		t=f.nextInt();
		precomp();
		PrintWriter out=new PrintWriter(System.out);
		while(t>0) {
			t--;
			int n=f.nextInt();
			int[] l=f.readArray(2*n);
			long sum=0;
			sort(l);
			for(int i=0;i<2*n;i++) {
				if(i<n) {
					sum=add(-l[i],sum);
				}
				else {
					sum=add(l[i],sum);
				}
				
			}
			
			System.out.println(mul(sum,nCk(2*n,n)));
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	