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 int INF = 1e9 + 7;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
const long long M = 998244353;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(2 * n + 1, 0), inv(n + 2, 0);
  inv[1] = 1;
  for (long long i = 1; i <= 2 * n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (long long i = 2; i <= n + 1; i++) {
    inv[i] = M - (M / i) * inv[M % i] % M;
  }
  long long ans = 1;
  for (long long i = 1; i <= 2 * n - 1; i++) {
    ans *= i;
    ans %= M;
  }
  for (long long i = 1; i <= n; i++) {
    ans *= inv[i];
    ans %= M;
  }
  for (long long i = 1; i <= n - 1; i++) {
    ans *= inv[i];
    ans %= M;
  }
  ans *= 2;
  ans %= M;
  long long temp = 0;
  for (long long i = n + 1; i <= 2 * n; i++) {
    temp = (temp + a[i]);
    temp %= M;
  }
  for (long long i = 1; i <= n; i++) {
    temp = (temp - a[i] + M);
    temp %= M;
  }
  cout << (ans * temp) % M << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    solve();
  }
  return 0;
}
