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). 
#include <bits/stdc++.h>
using namespace std;
long long mo = 998244353;
long long bin_fun(long long chto, long long vkak) {
  if (vkak == 0) return 1;
  if (vkak == 1) return chto;
  if (vkak % 2 == 0)
    return bin_fun(chto * chto % mo, vkak / 2);
  else
    return (bin_fun(chto * chto % mo, vkak / 2) * chto) % mo;
}
int main() {
  long long a, s, d, i, j, k;
  cin >> a;
  vector<long long> v1;
  for (i = 0; i < a * 2; i++) {
    cin >> s;
    v1.push_back(s);
  }
  sort(v1.begin(), v1.end());
  long long sum = 0;
  for (i = 0; i < a; i++) sum = (sum + abs(v1[a * 2 - i - 1] - v1[i])) % mo;
  long long ch1 = 1, ch2 = 1;
  for (i = a + 1; i <= a * 2; i++) {
    ch1 = (ch1 * i) % mo;
  }
  for (i = 1; i <= a; i++) {
    long long t = bin_fun(i, mo - 2);
    ch2 = (ch2 * t) % mo;
  }
  long long otv = (ch1 * ch2) % mo;
  cout << (otv * sum) % mo;
}
