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;
int const mod = 998244353;
int const maxn = 3e5 + 1998;
int a[maxn];
int f(int x) {
  int s = x;
  for (int i = 2; i * 1LL * i <= x; ++i)
    if (x % i == 0) {
      while (x % i == 0) x /= i;
      s -= s / i;
    }
  if (x > 1) s -= s / x;
  return s;
}
int modpow(int a, int b) {
  if (!b) return 1;
  if (b & 1) return a * 1LL * modpow(a, b - 1) % mod;
  int res = modpow(a, b >> 1);
  return res * 1LL * res % mod;
}
int n, n1;
long long s, p = 1, p1 = 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  n1 = 2 * n;
  for (int i = 0; i < n1; i++) cin >> a[i];
  sort(a, a + n1);
  for (int i = 0; i < n; i++) {
    (s += a[i + n] - a[i]) %= mod;
    (p *= i + 1) %= mod;
    (p1 *= i + n + 1) %= mod;
  }
  (p1 *= modpow(p, f(mod) - 1)) %= mod;
  cout << s * p1 % mod;
}
