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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return {i, j};
}
struct printer {
  ~printer() { cerr << endl; }
  template <class c>
  printer& operator,(c x) {
    cerr << boolalpha << x;
    return *this;
  }
  printer& operator,(string x) {
    cerr << x;
    return *this;
  }
  template <class c, class d>
  printer& operator,(pair<c, d> x) {
    return *this, "(", x.first, ", ", x.second, ")";
  }
  template <class... d, template <class...> class c>
  printer& operator,(c<d...> x) {
    return *this, range(begin(x), end(x));
  }
  template <class c>
  printer& operator,(rge<c> x) {
    *this, "[";
    for (auto it = x.b; it != x.e; ++it) *this, (it == x.b ? "" : ", "), *it;
    return *this, "]";
  }
};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long my_rand(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
const long long mod = 998244353;
const long long inf = -1;
const long long sz = 3e5 + 10;
long long ar[sz], fact[sz];
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < sz; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
}
long long bigmod(long long a, long long b, long long m) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2 == 1) x = (x * a) % m;
  return x;
}
long long ncr(long long n, long long r) {
  long long up = fact[n];
  long long down = (fact[n - r] * fact[r]) % mod;
  long long res = (up * bigmod(down, mod - 2, mod)) % mod;
  return res;
}
signed main() {
  long long n, m, i;
  pre();
  cin >> n;
  m = 2 * n;
  for (i = 0; i < m; i++) {
    cin >> ar[i];
  }
  sort(ar, ar + m);
  long long ans = 0;
  for (i = 0; i < n; i++) {
    ans = (ans + (ar[n + i] - ar[i])) % mod;
  }
  ans = (ans * ncr(m, n)) % mod;
  cout << ans << endl;
}
