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;
constexpr long long N = 1e6 + 100;
constexpr long long MOD = 998244353;
long long fac[N], inv[N];
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = 1LL * res * a % MOD;
    a = 1LL * a * a % MOD;
    b >>= 1;
  }
  return res;
}
long long C(long long n, long long m) {
  if (m < 0 or m > n) return 0;
  return 1LL * fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
void Init() {
  fac[0] = 1;
  for (long long i = 1; i < N; i++) fac[i] = 1LL * fac[i - 1] * i % MOD;
  inv[N - 1] = binpow(fac[N - 1], MOD - 2);
  for (long long i = N - 2; i >= 0; i--)
    inv[i] = 1LL * inv[i + 1] * (i + 1) % MOD;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  Init();
  long long n;
  cin >> n;
  n *= 2;
  vector<long long> a(n);
  for (auto &x : a) cin >> x;
  sort(a.begin(), a.end());
  vector<long long> b(a.rbegin(), a.rend());
  long long sum = 0;
  for (long long i = 0; i < n / 2; i++) {
    sum += abs(a[i] - b[i]);
    sum %= MOD;
  }
  cout << C(n, n / 2) * sum % MOD << "\n";
  return 0;
}
