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>
const int SIZE = 1e5;
using namespace std;
long long int mod_pow(long long int a, long long int x, long long int p) {
  long long int result = 1;
  while (x > 0) {
    if (x % 2 != 0) result = (result * a) % p;
    a = (a * a) % p;
    x /= 2;
  }
  return result;
}
long long int mod_inverse(long long int a, long long int p) {
  return mod_pow(a, p - 2, p);
}
long long int nCk(int n, int k, int p = 998244353) {
  long long int numerator = 1, denominator = 1;
  for (size_t i = 0; i < k; i++) numerator = (numerator * (n - i)) % p;
  for (size_t i = 1; i <= k; i++) denominator = (denominator * i) % p;
  return (numerator * mod_inverse(denominator, p)) % p;
}
long long int d(int p = 998244353) {
  int n;
  long long int left_sum = 0, right_sum = 0;
  cin >> n;
  vector<int> arr(2 * n), a(n), b(n);
  for (size_t i = 0; i < 2 * n; i++) cin >> arr[i];
  sort(arr.begin(), arr.end());
  for (size_t i = 0; i < n; i++) left_sum = (left_sum + arr[i]) % p;
  for (size_t i = n; i < 2 * n; i++) right_sum = (right_sum + arr[i]) % p;
  long long int result = (right_sum - left_sum) % p;
  if (result < 0) result += p;
  result = (result * nCk(2 * n, n)) % p;
  return result;
}
int main(int argc, char const *argv[]) {
  int t = 1;
  while (t--) {
    cout << d();
    cout << endl;
  }
  return 0;
}
