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;
int _____asdfwqerqwdcasdfasd = 0;
const long long mod = 998244353;
long long fact(long long x) {
  long long ans = 1;
  for (long long i = 2; i <= x; i++) {
    (ans *= i) %= mod;
  }
  return ans;
}
long long binpow(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ans = binpow(x, y / 2);
  (ans *= ans) %= mod;
  if (y % 2 == 1) {
    (ans *= x) %= mod;
  }
  return ans;
}
long long inv(long long x) { return binpow(fact(x), mod - 2); }
long long C(long long n, long long k) {
  return fact(n) % mod * inv(k) % mod * inv(n - k) % mod;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(2 * n);
  vector<long long> sm(2 * n);
  for (long long& x : a) {
    cin >> x;
  }
  sort(a.begin(), a.end());
  sm[0] = a[0];
  for (long long i = 1; i < 2 * n; i++) {
    sm[i] = sm[i - 1] + a[i];
  }
  long long x = sm[2 * n - 1] - 2 * sm[n - 1];
  long long ans = (((2ll * x) % mod) * C(2 * n - 1, n - 1) % mod) % mod;
  cout << ans << endl;
  return 0;
}
