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;
long long poww(long long x, long long y, long long m) {
  x = x % m;
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return poww(x * x, y / 2, m);
  } else {
    return (x * poww(x * x, y / 2, m)) % m;
  }
}
void io();
const int mod = 998244353;
const int maxn = 300000 + 10;
int n;
long long a[maxn];
long long fact[maxn];
int main() {
  io();
  int i, j;
  fact[0] = 1;
  cin >> n;
  for (long long i = 1; i <= 2 * n; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  for ((i) = 0; (i) < (2 * n); i = i + 1) cin >> a[i];
  sort(a, a + 2 * n);
  long long second = 0;
  for ((i) = n; (i) < (2 * n); i = i + 1) second += a[i];
  for ((i) = 0; (i) < (n); i = i + 1) second -= a[i];
  second %= mod;
  long long c1 = fact[2 * n];
  long long c2 = (fact[n] * fact[n]) % mod;
  c2 = poww(c2, mod - 2, mod) % mod;
  long long c3 = (c1 * c2) % mod;
  long long ans = (c3 * second) % mod;
  cout << ans;
  return 0;
}
void io() {}
