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.*;

public class A {
    static int mod = 998244353;
    static int fact[];

    static int size = 1000005;

    static long ncr(int n, int r) {
        fact = new int[size];
        fact[0] = 1;
        fact[1] = 1;
        for (int i = 2; i < size; i++) {
            fact[i] = (int) ((i * (long) fact[i - 1]) % mod);
        }
        if (r > n)
            return 0;
        return (long) fact[n] * power(fact[r], mod - 2) % mod * power(fact[n - r], mod - 2) % mod;
    }

    static long power(long a, long b) {
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1)
                res = res * a % mod;
            b >>= 1;
            a = a * a % mod;
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        int arr[] = new int[2 * n];

        for (int i = 0; i < 2 * n; i++) {
            arr[i] = sc.nextInt();
        }

        Arrays.sort(arr);

        long small_sum = 0;
        long large_sum = 0;

        for (int i = 0; i < 2 * n; i++) {
            if (i < n) {
                small_sum = (small_sum + arr[i]) % mod;
            } else {
                large_sum = (large_sum + arr[i]) % mod;
            }
        }

        long permutation = ncr(2 * n, n);

        long ans = (large_sum - small_sum + mod) % mod * permutation % mod;

        System.out.println(ans);

    }
}