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 long long mod = 998244353;
int a[411100], b[411100], pos[411100];
long long A[411100], A2[411100];
long long pw(long long a, int b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) {
      ret = (ret * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
long long c(int n, int m) {
  long long t1 = A[n];
  long long t2 = A[m];
  long long t3 = A[n - m];
  t1 = (t1 * pw(t2, mod - 2)) % mod;
  t1 = (t1 * pw(t3, mod - 2)) % mod;
  return t1;
}
int main() {
  int n;
  cin >> n;
  map<int, int> num;
  vector<int> ft;
  for (int i = 0; i < 2 * n; ++i) {
    cin >> a[i];
    num[a[i]]++;
    ft.push_back(a[i]);
  }
  sort(a, a + 2 * n);
  A2[0] = 1;
  A[0] = 1;
  for (int i = 1; i <= 2 * n; ++i) {
    A2[i] = (A2[i - 1] * 2) % mod;
    A[i] = (A[i - 1] * i) % mod;
  }
  long long tot = c(2 * n, n);
  long long ans = 0;
  for (int i = 0; i < 2 * n; ++i) {
    long long x = a[i];
    if (i < n) {
      ans = (ans - (tot * x) % mod + mod) % mod;
    } else {
      ans = (ans + (tot * x) % mod + mod) % mod;
    }
  }
  cout << ans << endl;
}
