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 F[300000 + 5];
void init() {
  F[0] = F[1] = 1;
  for (int i = 2; i <= 300000; i++) F[i] = (1LL * F[i - 1] * i) % 998244353;
}
long long power(long long A, long long N) {
  long long res = 1;
  while (N) {
    if (N & 1)
      res = (1LL * res * A) % 998244353, N--;
    else
      A = (1LL * A * A) % 998244353, N >>= 1;
  }
  return res;
}
long long nCr(long long n, long long k) {
  if (k > n) return 0;
  init();
  long long res = F[n];
  res = (1LL * res * power(F[k], 998244353 - 2)) % 998244353;
  res = (1LL * res * power(F[n - k], 998244353 - 2)) % 998244353;
  return res;
}
void solve() {
  int n;
  cin >> n;
  int m = 2 * n;
  long long arr[m];
  for (int i = 0; i < m; i++) cin >> arr[i];
  sort(arr, arr + m);
  long long ans = 0;
  for (int i = m - 1; i >= n; i--) ans = (ans + arr[i]) % 998244353;
  for (int i = n - 1; i >= 0; i--) ans = (ans - arr[i] + 998244353) % 998244353;
  ans = (1LL * ans * nCr(m, n)) % 998244353;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  solve();
  return 0;
}
