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;
void solve();
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}
long long modular_exp(long long a, long long n) {
  if (n == 0) return 1;
  long long res = 1;
  while (n > 0) {
    if (n % 2 != 0) res *= a;
    a = (a * a) % 998244353;
    n = n / 2;
    res = res % 998244353;
  }
  res = (res % 998244353);
  return res;
}
long long factorial[1500001];
void StoreFactorials(long long n) {
  factorial[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factorial[i] = (i * factorial[i - 1]) % 998244353;
  }
}
long long invmod(long long x) { return modular_exp(x, 998244353 - 2); }
long long nCr(long long n, long long r) {
  return (factorial[n] *
          invmod((factorial[r] * factorial[n - r]) % 998244353)) %
         998244353;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[2 * n];
  for (auto &x : arr) cin >> x;
  sort(arr, arr + 2 * n);
  long long res = 0;
  for (long long i = 0; i < 2 * n; i++) {
    if (i < n)
      res -= arr[i];
    else
      res += arr[i];
  }
  res %= 998244353;
  StoreFactorials(2 * n);
  res *= nCr(2 * n, n);
  res %= 998244353;
  cout << res;
  cout << "\n";
  return;
}
