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;
void fun() {}
long long int md = 998244353;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b, long long int md) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a) % md;
    }
    a = (1ll * a * a) % md;
    b >>= 1;
  }
  return res;
}
long long int poww(long long int a, long long int b) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a);
    }
    a = (1ll * a * a);
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
}
long long int divide(long long int a, long long int b, long long int md) {
  long long int rr = a * (poww(b, md - 2, md));
  rr %= md;
  return rr;
}
const long long int size = 3e5 + 5;
long long int n, ans;
long long int arr[size];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fun();
  cin >> n;
  for (long long int i = 1; i <= 2 * n; i++) {
    cin >> arr[i];
  }
  sort(arr + 1, arr + 1 + 2 * n);
  long long int ncr = 1, den = 1;
  for (long long int i = 1; i <= 2 * n; i++) ncr = (ncr * i) % md;
  for (long long int i = 1; i <= n; i++) den = den * i % md;
  ncr = divide(ncr, den, md);
  ncr = divide(ncr, den, md);
  long long int sum = 0;
  for (long long int i = 1; i <= 2 * n; i++) {
    if (i <= n)
      sum -= arr[i];
    else
      sum += arr[i];
    sum = (sum + md) % md;
  }
  ans = ncr * (sum % md) % md;
  cout << ans << '\n';
  return 0;
}
