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>
int main() {
  static const int MOD = 998244353;
  int n;
  std::cin >> n;
  std::vector<int> a(n * 2);
  std::vector<int64_t> fact(n * 2 + 5);
  std::vector<int64_t> inv(n * 2 + 5);
  std::vector<int64_t> inv_fact(n * 2 + 5);
  fact[0] = inv[0] = inv_fact[0] = 1;
  fact[1] = inv[1] = inv_fact[1] = 1;
  for (int i = 2; i <= n * 2; ++i) {
    fact[i] = (fact[i - 1] * i) % MOD;
    inv[i] = (inv[MOD % i] * (MOD - MOD / i)) % MOD;
    inv_fact[i] = (inv_fact[i - 1] * inv[i]) % MOD;
  }
  auto C = [&](int k, int n) {
    return (((fact[n] * inv_fact[n - k]) % MOD) * inv_fact[k]) % MOD;
  };
  for (int i = 0; i < 2 * n; ++i) {
    std::cin >> a[i];
  }
  std::sort(a.begin(), a.end());
  int64_t sum = 0;
  for (int i = 0; i < n * 2; ++i) {
    sum += i < n ? -a[i] : a[i];
  }
  sum = ((sum % MOD) + MOD) % MOD;
  std::cout << ((sum * C(n, n * 2)) % MOD);
}
