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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class b {
    static int n;

    static long[] fact, invFact;
    static final long INF_LONG = (long)(1e18 + 5), MOD = 998244353;

    public static void main(String[] args) {
        FastScanner fs=new FastScanner();
//        int T=fs.nextInt();


//        for (int tt=0; tt<T; tt++) {

            n = fs.nextInt();
            int [] a = fs.readArray(2 * n);
            sort(a);
            preCompute();

            long ans = 0;
        for(int i = 0; i < n; i++) ans -= a[i];
        for(int i = n; i < 2 * n; i++) ans += a[i];
            ans = ans % MOD;
            ans *= ncr(2 * n, n);
            ans = ans % MOD;

            System.out.println(ans);
        }







//            }






    static long power(long a, long b) {
        if(b == 0) return 1l;
        long temp = power(a, b / 2);
        temp = (temp * temp) % MOD;
        return b % 2 == 0 ? temp : (temp * a) % MOD;
    }

    static void preCompute() {
        fact = new long[2 * n + 5];
        invFact = new long[2 * n + 5];
        fact[0] = 1l;
        for(int i = 1; i <= 2 * n; i++) fact[i] = (fact[i - 1] * i) % MOD;
        invFact[2 * n] = power(fact[2 * n], MOD - 2);
        for(int i = 2 * n - 1; i >= 0; i--) invFact[i] = (invFact[i + 1] * (i + 1)) % MOD;
    }


    private static long ncr(int n, int c) {
        return (((fact[n] * invFact[c]) % MOD) * invFact[n - c]) % MOD;
    }


    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.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());
        }
    }


}

