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 N = 1e6 + 5;
const long long inf = 1e18 + 5;
void fileIO();
template <class T>
void print_container(T &container);
template <class T>
void print_pair_container(T &container);
template <class T>
void printArr(T &a, long long n);
template <typename T>
void debug(string s, T x);
template <typename T, typename... Args>
void debug(string s, T x, Args... args);
long long n, m;
long long a[N];
long long fact[N], invfact[N];
long long power(long long a, long long n, long long m) {
  a %= m;
  long long res = 1;
  while (n > 0) {
    if (n % 2) res = res * a % m;
    a = a * a % m;
    n /= 2;
  }
  return res;
}
long long modinv(long long n, long long m = 998244353) {
  return power(n, m - 2, m);
}
long long ncr(long long n, long long r) {
  if (r > n || n < 0 || r < 0) {
    return 0;
  } else {
    long long val =
        fact[n] * invfact[r] % 998244353 * invfact[n - r] % 998244353;
    return val;
  }
}
void precompute() {
  fact[0] = fact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
    fact[i] %= 998244353;
  }
  invfact[N - 1] = modinv(fact[N - 1]);
  for (long long i = N - 2; i >= 0; i--) {
    invfact[i] = invfact[i + 1] * (i + 1);
    invfact[i] %= 998244353;
  }
}
void solve() {
  cin >> n;
  for (long long i = 0; i < 2 * n; i++) {
    cin >> a[i];
  }
  sort(a, a + 2 * n);
  long long val = 0;
  for (long long i = 0; i < n; i++) {
    val += abs(a[i] - a[2 * n - i - 1]);
    val %= 998244353;
  }
  long long nr = ncr(2 * n, n);
  cout << (nr * val) % 998244353 << "\n";
}
int32_t main() {
  fileIO();
  precompute();
  solve();
  return 0;
}
void fileIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
template <typename T>
void debug(string s, T x) {
  cout << s << " = " << x << "\n";
}
template <typename T, typename... Args>
void debug(string s, T x, Args... args) {
  cout << s.substr(0, s.find(',')) << " = " << x << " | ";
  debug(s.substr(s.find(',') + 2), args...);
}
template <class T>
void print_container(T &container) {
  for (auto &t : container) {
    cout << t << " ";
  }
  cout << "\n";
}
template <class T>
void print_pair_container(T &container) {
  for (auto &t : container) {
    cout << t.first << " " << t.second << "\n";
  }
  cout << "\n";
}
template <class T>
void printArr(T &a, long long n) {
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
