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 M = 998244353;
long long fastpower(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return fastpower((x * x) % M, n / 2, M);
  else
    return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool vowl(char c) {
  return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
long long modInverse(long long A, long long M) {
  return fastpower(A, M - 2, M);
}
void sieve(long long N) {
  bool isPrime[N + 1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
vector<long long> factorize(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long m = 2 * n;
  long long arr[m];
  for (long long i = 0; i < m; i++) cin >> arr[i];
  sort(arr, arr + m);
  long long fact[300005];
  fact[0] = 1;
  for (long long i = 1; i <= 300000; i++) fact[i] = (fact[i - 1] * i) % M;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += arr[m - i - 1] - arr[i];
    ans %= M;
  }
  ans *= fact[m];
  ans %= M;
  ans *= fastpower(fact[n], M - 2, M);
  ans %= M;
  ans *= fastpower(fact[n], M - 2, M);
  ans %= M;
  cout << ans << endl;
  return 0;
}
