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 mod = 998244353;
long long expo_pow(long long x, long long y) {
  if (y == 0) return 1;
  y = y % (mod - 1);
  x %= mod;
  if (y == 0) y = mod - 1;
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
long long add() { return 0; }
template <typename T, typename... Types>
T add(T var1, Types... var2) {
  return (((((long long)(var1)) % mod + (long long)(add(var2...))) % mod) +
          mod) %
         mod;
}
long long mul() { return 1; }
template <typename T, typename... Types>
T mul(T var1, Types... var2) {
  return (((long long)(var1)) % mod * (long long)(mul(var2...))) % mod;
}
const long long mx = 3e5;
long long fact[mx];
long long inv_fact[mx];
long long inv(long long x) { return expo_pow(x, mod - 2); }
void compute() {
  fact[0] = 1;
  for (long long i = 1; i < mx; ++i) fact[i] = mul(fact[i - 1], i);
  for (long long i = 0; i < mx; ++i) inv_fact[i] = inv(fact[i]);
}
void solve() {
  compute();
  long long n;
  cin >> n;
  long long a[2 * n];
  for (long long i = 0; i < 2 * n; ++i) cin >> a[i];
  sort(a, a + 2 * n);
  long long tmp = mul(fact[2 * n - 1], inv_fact[n], inv_fact[n - 1]);
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    ans = add(ans, a[2 * n - 1 - i] - a[i]);
  }
  ans = mul(ans, tmp, 2);
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  long long i = 1;
  while (test--) {
    solve();
  }
}
