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.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B {

    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    private static StringTokenizer st;

    static long MOD = 998244353;

    public static void main(String[] args) throws Exception {
        st = new StringTokenizer(reader.readLine());
        int n = Integer.parseInt(st.nextToken());
        calculate_factorial(2 * n, MOD);
        long[] a = new long[2 * n];
        st = new StringTokenizer(reader.readLine());
        for (int i = 0; i < 2 * n; i++) {
            a[i] = Long.parseLong(st.nextToken());
        }
        Arrays.sort(a);
        long oneCost = 0;
        for (int i = 0; i < n; i++) {
            oneCost += Math.abs(a[2 * n - 1 - i] - a[i]);
        }
        oneCost %= MOD;
        writer.write(Long.toString(oneCost * choose(2 * n, n, MOD) % MOD));
        writer.flush();
    }

    public static long inverse(long a, long n) {
        long t = 0;
        long newt = 1;
        long r = n;
        long newr = a;
        while (newr != 0) {
            long quotient = r / newr;
            long newt_temp = newt;
            newt = t - quotient * newt;
            t = newt_temp;
            long newr_temp = newr;
            newr = r - quotient * newr;
            r = newr_temp;
        }
        if (r > 1) {
            // a is not invertible
            return -1;
        }
        if (t < 0) {
            t += n;
        }
        return t;
    }

    static long[] factorial;

    static void calculate_factorial(int N, long p) {
        factorial = new long[N + 1];
        factorial[0] = 1;
        for (int i = 1; i <= N; i++) {
            factorial[i] = factorial[i - 1] * i % p;
        }
    }

    static long choose(int n, int k, long p) {
        return factorial[n] * inverse(factorial[k] * factorial[n - k] % p, p) % p;
    }

}
