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  D{
	
	static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        private FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

	 int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
	}

	int[] nextIntArrayOne(int n) {
				int[] a = new int[n+1];
				for (int i = 1; i < n+1; i++)
					a[i] = nextInt();
				return a;
	}

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
	

	
	
	static long mod = (int) (998244353);
	static int N = 1000000;
	static long fact[] = new long[N];
	static long invfact[] = new long[N];

	static int power(long a, long n, long p) {
		long result = 1;
		while (n > 0) {
			if ((n & 1) > 0)
				result = a * result % p;
			a = a * a % p;
			n >>= 1;
		}
		return (int) result;
	}

	static void init() {
		long p = mod;
		fact[0] = 1;
		int i;
		for (i = 1; i < N; i++) {
			fact[i] = i * fact[i - 1] % p;
		}
		i--;
		invfact[i] = power(fact[i], p - 2, p);
		for (i--; i >= 0; i--) {
			invfact[i] = invfact[i + 1] * (i + 1) % p;
		}
	}

	static int ncr(int n, int r) {
		if (r > n || n < 0 || r < 0)
			return 0;
		return (int) ((int) (fact[n] * invfact[r] % mod * invfact[n - r] % mod) % mod);
	}
    
	public static void main(String[] args) {
    	FastReader s = new FastReader();
    	StringBuilder str = new StringBuilder();
    	
    	init();
		int n = s.nextInt();
		int arr[] = s.nextIntArray(2 * n);
		
		Arrays.sort(arr);
		
		long smallSum = 0;
		
		for(int i = 0; i < n; i++) {
			smallSum = (smallSum % mod  + arr[i] % mod) % mod;
		}
		
		long largeSum = 0;
		
		for(int i = n; i < 2 * n; i++) {
			largeSum  = (largeSum % mod  + arr[i] % mod) % mod;;
		}
		
		long cof = ncr(2 * n, n);
		
		long ans = (largeSum % mod - smallSum % mod + mod) % mod;
		
		ans = ((ans % mod) * (cof % mod)) % mod;
		
		System.out.println(ans);
     }

	



	

}

