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;
using s8 = char;
using u8 = unsigned char;
using s16 = short;
using u16 = unsigned short;
using s32 = int;
using u32 = unsigned int;
using s64 = long long;
using u64 = unsigned long long;
using ld = long double;
const s64 mod = 998244353;
s64 epow(s64 a, s64 n) {
  if (n == 0) {
    return 1;
  } else {
    s64 z = epow(a, n / 2);
    z = (z * z) % mod;
    if (n % 2 == 1) {
      z = (z * a) % mod;
    }
    return z;
  }
}
s64 fact(s64 n) {
  s64 res = 1;
  while (n) {
    res = (res * n) % mod;
    n--;
  }
  return res;
}
s64 inv(s64 x) { return epow(x, mod - 2); }
s64 cnk(s64 n, s64 k) {
  return (fact(n) * inv((fact(k) * fact(n - k)) % mod)) % mod;
}
void solve(istream& cin) {
  int n;
  cin >> n;
  n *= 2;
  vector<s64> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  s64 cost = 0;
  for (int i = 0; i < n / 2; i++) {
    cost = (cost + abs(a[i] - a[i + n / 2])) % mod;
  }
  s64 res = (cost * cnk(n, n / 2)) % mod;
  cout << res << "\n";
}
int main() {
  ios::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  int t = 1;
  while (t--) {
    solve(cin);
  }
  return 0;
}
