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 int64_t Mod = 998244353;
int64_t fac[300008], ifac[300008];
int64_t pw(int64_t a, int64_t b) {
  int64_t res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % Mod;
    a = 1ll * a * a % Mod;
    b /= 2;
  }
  return res;
}
int64_t C(int64_t n, int64_t k) {
  if (k > n || k < 0) return 0;
  return 1ll * fac[n] * ifac[k] % Mod * ifac[n - k] % Mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int64_t i = fac[0] = ifac[0] = 1; i < 300008; i++) {
    fac[i] = 1ll * fac[i - 1] * i % Mod;
    ifac[i] = pw(fac[i], Mod - 2);
  }
  int64_t n;
  cin >> n;
  vector<int64_t> a(2 * n);
  int64_t sum1 = 0, sum2 = 0, ans = 0;
  for (int64_t i = 0; i < 2 * n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (int64_t i = 0; i < n; ++i) {
    sum1 += a[i];
    sum1 %= Mod;
  }
  for (int64_t i = n; i < 2 * n; ++i) {
    sum2 += a[i];
    sum2 %= Mod;
  }
  ans = sum2 - sum1;
  if (ans < 0) ans += Mod;
  int64_t c = C(2 * n, n);
  cout << (ans * c) % Mod << endl;
  return 0;
}
