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 std::cerr;
using std::cin;
using std::max;
using std::min;
int n, A[300005], fac[300005], inv[300005], ans;
int cal(int x) {
  x -= x < 998244353 ? 0 : 998244353;
  return x;
}
int C(int x, int y) {
  return x < 0 || y < 0 || x < y
             ? 0
             : (long long)fac[x] * inv[y] % 998244353 * inv[x - y] % 998244353;
}
int main() {
  scanf("%d", &n), n <<= 1, fac[0] = inv[0] = inv[1] = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % 998244353;
  for (int i = 2; i <= n; i++)
    inv[i] =
        (long long)(998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  for (int i = 2; i <= n; i++)
    inv[i] = (long long)inv[i - 1] * inv[i] % 998244353;
  std::sort(A + 1, A + 1 + n);
  for (int i = 1; i <= n; i++)
    ans = i > n / 2 ? cal(ans + A[i] % 998244353)
                    : cal(ans - A[i] % 998244353 + 998244353);
  ans = 2ll * ans * C(n - 1, n / 2 - 1) % 998244353;
  printf("%d\n", ans);
  return 0;
}
