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 int mar = 998244353;
void fastpow(long long a, long long b, long long &bp) {
  vector<int> logp;
  while (b > 0) {
    if (b % 2 == 1) {
      logp.push_back(1);
    } else {
      logp.push_back(0);
    }
    b = b / 2;
  }
  for (int i = 0; i < logp.size(); i++) {
    long long ujb = a;
    if (logp[i] == 1) {
      for (int j = 0; j < i; j++) {
        ujb = ujb * ujb;
        ujb = ujb % 998244353;
      }
      bp = bp * ujb;
      bp = bp % 998244353;
    }
  }
}
void binomial(long long n, long long k, long long &bi, long long p) {
  k = min(k, n - k);
  long long sz = 1, ne = 1;
  for (int i = n; i > n - k; i--) {
    sz = sz * i;
    sz = sz % p;
  }
  for (int i = k; i > 1; i--) {
    ne = ne * i;
    ne = ne % p;
  }
  fastpow(ne, p - 2, bi);
  bi = bi * sz;
  bi = bi % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  priority_queue<int> a;
  for (int i = 0; i < n * 2; i++) {
    int x;
    cin >> x;
    a.push(x);
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    sum += a.top();
    sum %= mar;
    a.pop();
  }
  for (int i = 0; i < n; i++) {
    sum -= a.top();
    sum += mar;
    sum %= mar;
    a.pop();
  }
  long long ki = 1;
  binomial(2 * n, n, ki, mar);
  ki %= mar;
  ki *= sum;
  ki %= mar;
  cout << ki << endl;
  return 0;
}
