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;
using ll = long long;
const int mod = 998244353, N = 3e5 + 10;
ll fa[N];
ll power(ll x, ll y) {
  ll res = 1;
  while (y) {
    if (y % 2) res = (res * x) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
ll inv(ll x) {
  x %= mod;
  return power(x, mod - 2);
}
ll c(ll n, ll k) { return ((fa[n] * inv(fa[k])) % mod * inv(fa[n - k])) % mod; }
int main() {
  fa[0] = 1;
  for (int i = 1; i < N; i++) fa[i] = (i * fa[i - 1]) % mod;
  int n;
  cin >> n;
  vector<int> v(2 * n);
  for (int i = 0; i < 2 * n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  ll f = 0, s = 0;
  for (int i = 0; i < 2 * n; i++) {
    if (i < n)
      f += v[i];
    else
      s += v[i];
  }
  ll ans = (s - f) % mod;
  ans *= c(2 * n, n);
  cout << (ans + mod) % mod;
  return 0;
}
