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). 
/*
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    .add(key, value)
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Main {

    static final long M = 1000000007;
    private int[] ndigits;

    // use main for only io
    public static void main(String args[]) {
        FastReader io = new FastReader();
        int n = io.nextInt();

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

        for(int i = 0;i < 2*n;i++)a[i] = io.nextInt();

        Solver solver = new Solver();

        System.out.println(solver.solve(n,a));

    }
}

class Pair{
    int x,y;

    public Pair(int x, int y){
        this.x = x;
        this.y = y;
    }
}


class Solver {
    static final int M = 998244353;

    long solve(int n, int a[]) {

        Arrays.sort(a);
        long sum = 0;

//        for(int i : a)System.out.println(i);

        for(int i = 0;i < n;i++){
            sum = (sum + Math.abs(a[i] - a[2*n-i-1]))%M;
        }
        long fact[] = new long[2*n + 1];

        factorial(fact, 2*n);

//        System.out.println(modexp(fact[n], M-2));

        long ans = (sum*((((fact[2*n]*inv(fact[n]))%M)*inv(fact[n]))%M))%M;
        return ans;
    }

    // returns the first key greater than or equal to val
    int lower_bound(int a[], int val) {
        int low = 0, high = a.length - 1, ret = -1;

        while (low <= high) {
            int mid = (low + high) / 2;

            if (a[mid] < val) low = mid + 1;
            else {
                ret = mid;
                high = mid - 1;
            }

        }

        return ret;
    }

    // returns the first key strictly greater than val
    int upper_bound(int a[], int val) {
        int low = 0, high = a.length - 1, ret = -1;

        while (low <= high) {
            int mid = (low + high) / 2;

            if (a[mid] <= val) low = mid + 1;
            else {
                ret = mid;
                high = mid - 1;
            }

        }

        return ret;
    }
    long modexp(long n, int m) {
        if (m == 0)
            return 1;
        else if (m == 1)
            return n;
        else {
            long p = modexp(n, m / 2);

            if (m % 2 == 1)
                return (((p * p) % M) * n) % M;
            else
                return (p * p) % M;
        }
    }

    long exp(long n, long m) {
        if (m == 0)
            return 1;
        if (m == 1)
            return n;

        long p = exp(n, m / 2);

        if (m % 2 == 1)
            return p * p * n;
        else
            return p * p;
    }

    long gcd(long a, long b) {
        if (a == 0)
            return b;

        return gcd(b % a, a);
    }

    long inv(long n) {
        return modexp(n, M - 2);
    }

    long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    long factorial(long fact[], int n) {
        fact[0] = 1;
        fact[1] = 1;

        long prod = 1;

        for (int i = 2; i <= n; i++) {
            prod = (prod * i) % M;

            fact[i] = prod;
        }

        return prod;
    }

    boolean isPrime(long n) {
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
}


class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public 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());
    }

    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;
    }
}