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.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Random;
import java.util.StringTokenizer;

public class Solution{
	
	static final int mod = 998244353;
	static final int N = 300005;
	
	
    public static void main(String[] args) throws IOException {
     
    	FastScanner fs = new FastScanner();
    	PrintWriter out = new PrintWriter(System.out);
    	
    		
    		
    	int tt = 1;
    	while(tt-->0) {
    		
    		int n = fs.nextInt();
    		int[] a = fs.readArray(2*n);
    		
    		int[] fact = new int[N];
    		fact[0] = 1;
    		for(int i=1;i<N;i++) fact[i] = mul(fact[i-1], i);
    		
    		
    		
    		ruffleSort(a);
    		
    		int num1 = 0, num2 = 0;
    		for(int i=0;i<n;i++) {
    			num1 = (num1+a[i])%mod;
    			num2 = (num2+a[i+n])%mod;
    		}
    		
    		int ans = dif(num2, num1);
    		
    		int f = mul(fact[2*n], inv(mul(fact[n], fact[n])));
    		
    		ans = mul(ans, f);
    		out.println(ans);
    		
    		
    		
    	}
    	
    		
    		
    		
    	out.close();
    		
    }
    
    
    static int dif(int a, int b) {
    	a -= b;
    	if(a<0) return a+mod;
    	return a;
    }
    
   
   
    static int add(int a, int b) {
    	a += b;
    	if(a>=mod) return a-mod;
    	return a;
    }
    
    static int mul(int a, int b) {
    	return (int)(((long)a*b)%mod);
    }
    
    
    
    static int exp(int a, int n) {
    	int res = 1;
    	while(n>0) {
    		if((n&1)==1) {
    			res = mul(res, a);
    		}
    		a = mul(a, a);
    		n = n>>1;
    	}
    	return res;
    }
    
    static int inv(int n) {
    	return exp(n, mod-2);
    }
    
  
    
    
    static final Random random=new Random();
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }
   	
}
