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;
long long int fact[300005];
long long int sum(long long int a, long long int b) {
  return ((a % 998244353) + (b % 998244353)) % 998244353;
}
long long int diff(long long int a, long long int b) {
  return (((a + 998244353) % 998244353 - (b + 998244353) % 998244353) +
          998244353) %
         998244353;
}
long long int mul(long long int a, long long int b) {
  return ((a % 998244353) * (b % 998244353)) % 998244353;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2 == 0) {
      a = mul(a, a);
      b /= 2;
    } else {
      ans = mul(ans, a);
      b--;
    }
  }
  return ans;
}
long long int NCR(long long int N, long long int R) {
  long long int t1 = fact[N];
  long long int t2 = power(fact[R], 998244353 - 2);
  long long int t3 = power(fact[N - R], 998244353 - 2);
  return mul(mul(t1, t2), t3);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  fact[0] = 1;
  for (int i = 1; i < 300005; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  int N;
  cin >> N;
  int a[2 * N], b[2 * N];
  for (int i = 0; i < 2 * N; i++) {
    cin >> a[i];
  }
  sort(a, a + 2 * N);
  long long int ans = 0, p = 0;
  ans = NCR(2 * N, N);
  for (int i = 0; i < N; i++) {
    p = sum(p, abs(a[i] - a[2 * N - (i + 1)]));
  }
  ans = mul(ans, p);
  cout << ans << '\n';
  return 0;
}
