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;
void no() { printf("NO\n"); }
void yes() { printf("YES\n"); }
void files() {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
  return;
}
const long long N = 1e5 + 5;
const long long M = 998244353;
long long binPow(long long base, long long pow) {
  if (base == 0 && M + 2 == M) return 1;
  base %= M;
  long long ans = 1;
  base = base % M;
  while (pow) {
    if (pow & 1) ans = ans * base % M;
    base = base * base % M;
    pow >>= 1;
  }
  return ans;
}
unsigned long long mulMod(unsigned long long a, unsigned long long b,
                          unsigned long long m) {
  long double x;
  unsigned long long c;
  long long r;
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  x = a;
  c = x * b / m;
  r = (long long)(a * b - c * m) % (long long)m;
  return r < 0 ? r + m : r;
}
long long exEuclid(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = exEuclid(b % a, a, x1, y1);
  x = y1 - b / a * x1;
  y = x1;
  return gcd;
}
void sol() {
  int n;
  cin >> n;
  vector<long long> v(2 * n);
  for (long long &in : v) cin >> in;
  sort(v.begin(), v.end());
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans = (ans + v[n + i] - v[i]) % M;
  }
  long long fact[2 * n + 1];
  for (long long i = fact[0] = 1; i <= 2 * n; ++i) {
    fact[i] = (fact[i - 1] * i) % M;
  }
  ans = ans * fact[2 * n] % M;
  ans = ans * binPow(fact[n], M - 2) % M;
  ans = ans * binPow(fact[n], M - 2) % M;
  cout << ans << "\n";
}
int main() {
  int tc = 1;
  while (tc--) {
    sol();
  }
}
