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 itertools
import math
import sys
import os
from collections import defaultdict
from heapq import heapify, heappush, heappop 
 
def is_debug():
    return "PYPY3_HOME" not in os.environ

def stdin_wrapper():
    data = '''5
13 8 35 94 9284 34 54 69 123 846

'''
    
    for line in data.split('\n'):
        yield line
 
if not is_debug():
    def stdin_wrapper():
        while True:
            yield sys.stdin.readline()
 
inputs = stdin_wrapper()
 
def input_wrapper():
    return next(inputs)
 


def get_str():
    if is_debug():
        return input_wrapper()

    return input()

def get(_type):
    if _type == str:
        return get_str()

    return _type(input_wrapper())
 
def get_arr(_type):
    return [_type(x) for x in input_wrapper().split()]
 

def tuplerize(method):
    def wrap(*args, **kwargs):
        res = method(*args, **kwargs)
        if not isinstance(res, (tuple, list)):
            res = (res, )

        return res
    return wrap

''' Solution '''

@tuplerize
def solve(n, d):
    MOD = 998244353 
    def price(a,b):
        return sum([abs(x-y) % MOD for x, y in zip(sorted(a), reversed(sorted(b)))]) % MOD

    _MAX = 2*n + 1

    fact = [0 for _ in range((_MAX))]
    fact[0] = 1
    fact_inv = [0 for _ in range((_MAX))]
    for i in range(1, _MAX):
        fact[i] = fact[i-1] * i % MOD

    def fastpow(x, y, z):
        "Calculate (x ** y) % z efficiently."
        number = 1
        while y:
            if y & 1:
                number = number * x % z
            y >>= 1
            x = x * x % z
        return number


    fact_inv[_MAX-1] = fastpow(fact[_MAX-1], MOD - 2, MOD) % MOD
    for i in reversed(range(_MAX-1)):
        fact_inv[i] = fact_inv[i+1] * (i+1) % MOD

    p = price(d[:len(d)//2], d[len(d)//2:])

    

    return (p * ((((fact[len(d)] * fact_inv[len(d)//2]) % MOD) * fact_inv[len(d)//2]) % MOD)) % MOD


n = get(int)
a = get_arr(int)
print(*solve(n, a))