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 MOD = 998244353;
const double eps = 1e-10;
const long long INF = 1e18;
const int N = 3e5 + 10;
inline int inv(int a) {
  if (a < 0) a += MOD;
  int b = MOD, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += MOD;
  return u;
}
int fact[N], invfact[N];
void init() {
  fact[0] = 1;
  int i;
  for (i = 1; i < N; i++) {
    fact[i] = (1ll * i * fact[i - 1]) % MOD;
  }
  i--;
  invfact[i] = inv(fact[i]);
  for (i--; i >= 0; i--) {
    invfact[i] = (1ll * invfact[i + 1] * (i + 1)) % MOD;
  }
}
int ncr(int n, int r) {
  return (((1ll * fact[n] * invfact[r]) % MOD) * 1ll * invfact[n - r]) % MOD;
}
void solve() {
  int n;
  cin >> n;
  vector<int> odd, even;
  for (int i = 0; i < 2 * n; i++) {
    int x;
    cin >> x;
    if (i & 1)
      odd.push_back(x);
    else
      even.push_back(x);
  }
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end(), greater<int>());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += abs(odd[i] - even[i]);
  }
  ans %= MOD;
  ans *= ncr(2 * n, n);
  cout << ans % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
