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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import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;

   public class D3
   {
      static final long MOD = 998244353L;
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int[] arr = new int[N*2];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N*2; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         N <<= 1;
         sort(arr);
         fac = new long[N+1];
         invfac = new long[N+1];
         fac[0] = invfac[0] = 1L;
         for(int i=1; i <= N; i++)
            fac[i] = (fac[i-1]*i)%MOD;
         invfac[N] = power(fac[N], MOD-2, MOD);
         for(int i=N-1; i >= 1; i--)
            invfac[i] = (invfac[i+1]*(i+1))%MOD;
         long sum = 0L;
         for(int i=0; i < N/2; i++)
            sum += arr[N/2+i]-arr[i];
         sum %= MOD;
         long res = (((sum*fac[N])%MOD)*invfac[N/2])%MOD;
         System.out.println((res*invfac[N/2])%MOD);
      }
      static long[] fac;
      static long[] invfac;
      public static long nCr(int a, int b)
      {
         long res = (fac[a]*invfac[b])%MOD;
         return (res*invfac[a-b])%MOD;
      }
      public static long power(long x, long y, long p) 
      { 
         //0^0 = 1
         long res = 1L;      
         x = x%p;  
         while(y > 0) 
         { 
            if((y&1)==1) 
               res = (res*x)%p;
            y >>= 1;
            x = (x*x)%p;  
         } 
         return res; 
      }
      public static void sort(int[] arr)
      {
         //stable heap sort
         PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
         for(int a: arr)
            pq.add(a);
         for(int i=0; i < arr.length; i++)
            arr[i] = pq.poll();
      }
   }