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>
#pragma GCC optimize(2)
using namespace std;
const long long mod = 998244353;
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
mt19937 myrand(seed);
uniform_int_distribution<long long> dist(0, 5);
long long read() {
  long long x = 0, flag = 0;
  char ch = getchar();
  if (ch == '-') flag = 1;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x *= 10, x += ch - '0', ch = getchar();
  if (flag) return -x;
  return x;
}
long long power(long long a, long long n) {
  long long sum = 1;
  while (n) {
    if (n & 1) sum *= a;
    sum %= mod;
    a *= a;
    a %= mod;
    n >>= 1;
  }
  return sum % mod;
  ;
}
long long ny(long long n) { return power(n, mod - 2); }
long long a[500010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  long long i;
  cin >> n;
  for (i = 1; i <= 2 * n; i++) cin >> a[i];
  sort(a + 1, a + 2 * n + 1);
  long long sum = 0;
  for (i = 1; i <= 2 * n; i++)
    if (i <= n)
      sum -= a[i];
    else
      sum += a[i];
  sum %= mod;
  for (i = n + 1; i <= 2 * n; i++) {
    sum *= i;
    sum %= mod;
  }
  for (i = 1; i <= n; i++) {
    sum *= ny(i);
    sum %= mod;
  }
  cout << sum;
}
