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). 
from sys import stdin, stdout

# observation:
#    A         B
# 1,2,3,4 | 5,6,7,8
# for any p, q, index i.
# if pi belongs to A, then qi belongs to B.
# if qi belongs to A, then pi belongs to B.
# prove (contradiction):
# p: ____ i _____
# q: ____ i _____
# if both pi and qi belongs to A
# then p[0 : i] belongs to A, q[i : n-1] belongs to A.
# A has at least n + 1 elements, contradiction.

MOD = 998244353


def divide_and_sum(n, a_a):
    a_a.sort()
    s = (sum(a_a[n:2*n]) - sum(a_a[0:n])) % MOD
    c = Com(n, 2*n)
    r = (s * c) % MOD
    return r


def Com(p, t):
    c1 = 1
    c2 = 1
    for i in range(p):
        c1 *= (t-i)
        c1 %= MOD
        c2 *= (i+1)
        c2 %= MOD

    # MOD: c1 / c2 = c1 * c2^-1
    c = c1 * inverse(c2)
    c %= MOD
    return c


def inverse(a):
    return bpower(a, MOD-2)


def bpower(a, p):
    if p == 0:
        return 1

    r = bpower(a, p // 2)
    r *= r
    r %= MOD

    if p % 2 == 1:
        r *= a
    r %= MOD
    return r


n = int(stdin.readline())
a_a = list(map(int, stdin.readline().split()))
r = divide_and_sum(n, a_a)
stdout.write(str(r))