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;
const long long inf = 1e18 + 5;
const int maxN = 1e5 + 1;
long long power(long long a, long long n) {
  if (n == 0) {
    return 1ll;
  }
  if (n == 1) {
    return a % 998244353;
  }
  long long p = power(a, n / 2);
  p = (p * p) % 998244353;
  if (n % 2) {
    p = (p * a) % 998244353;
  }
  return p;
}
int main() {
  int n;
  cin >> n;
  vector<long long> a(n * 2);
  for (int i = 0; i < n * 2; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long long sum = 0ll;
  for (int i = 0; i < n * 2; i++) {
    if (i < n) {
      sum -= a[i];
    } else {
      sum += a[i];
    }
  }
  sum = (sum % 998244353);
  long long comb = 1ll;
  for (long long i = 0; i < n; i++) {
    comb = (((comb * (n * 2 - i)) % 998244353) * power(i + 1, 998244353 - 2)) %
           998244353;
  }
  long long ans = ((comb * sum) % 998244353 + 998244353) % 998244353;
  cout << ans << endl;
}
