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 sys
import math
from collections import defaultdict,Counter

# input=sys.stdin.readline
# def print(x):
#     sys.stdout.write(str(x)+"\n")

# sys.stdout=open("CP2/output.txt",'w')
# sys.stdin=open("CP2/input.txt",'r')

def power(x,y,m):
    if y==0:
        return 1
    p=power(x,y//2,m)%m
    p=(p*p)%m
    if y%2==0:
        return p
    else:
        return((x*p)%m)

def find(n):
	p=1
	q=1
	c=1
	while c<=n//2:
		p=(p*(n-c+1))%mod
		q=(q*c)%mod
		c+=1
	return (p*power(q,mod-2,mod))%mod
# mod=pow(10,9)+7
mod=998244353
n=int(input())
a=list(map(int,input().split()))
a.sort()
# print(a)
ans=0
for i in range(n):
	ans=(ans+a[i+n]-a[i])%mod
# ans=(ans*2)%mod
# print(ans)
mul=find(2*n)
print((ans*mul)%mod)
# for a1,b1 in zip(a,reversed(a)):
# 	ans+=abs(a1-b1)
# print(2588544/ans)
# print(2588544/252)
