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;
using P = pair<int, int>;
const int M = 998244353;
ll add(ll a, ll b) { return (a + b) % M; }
ll mul(ll a, ll b) { return (a * b) % M; }
ll mod_pow(ll a, ll b) {
  ll r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    b >>= 1;
    a = mul(a, a);
  }
  return r;
}
struct combination {
  vector<int> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < M);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = mul(fact[i - 1], i);
    ifact[n] = mod_pow(fact[n], M - 2);
    for (int i = n; i >= 1; --i) ifact[i - 1] = mul(ifact[i], i);
  }
  int operator()(int n, int k) {
    assert(k >= 0 && k <= n);
    return mul(fact[n], mul(ifact[k], ifact[n - k]));
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(2 * n);
  for (int i = 0; i < (int)(2 * n); ++i) cin >> a[i];
  sort(a.begin(), a.end());
  combination C(2 * n + 5);
  ll ans = 0;
  for (int i = 0; i < (int)(n); ++i) {
    ans = add(ans, M - a[i]);
    ans = add(ans, a[n + i]);
  }
  ans = mul(ans, C(2 * n, n));
  cout << ans << endl;
}
