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 INF = 1000000000 + 5;
const int N = 4e5 + 5;
const int M = 131072;
const long long oo = LLONG_MAX;
const int mod = 998244353;
void add_self(long long& a, long long b) {
  a += b;
  a %= mod;
}
long long mul(long long a, long long b) { return 1ll * a * b % mod; }
long long poww(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
long long inv(long long x) { return poww(x, mod - 2); }
long long fac[N];
int bino(long long n, long long k) {
  return mul(fac[n], mul(inv(fac[k]), inv(fac[n - k])));
}
long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
void init() {
  fac[0] = fac[1] = 1;
  for (int i = 2; i < N; ++i) fac[i] = mul(fac[i - 1], i);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  init();
  int n;
  cin >> n;
  vector<long long> a(n * 2);
  for (int i = 0; i < n * 2; ++i) cin >> a[i];
  sort((a).begin(), (a).end());
  long long sum1 = 0, sum2 = 0;
  for (int i = 0; i < n; ++i) add_self(sum1, a[i]);
  for (int i = n; i < n * 2; ++i) add_self(sum2, a[i]);
  long long diff = 0;
  add_self(diff, add(sum2, -sum1));
  cout << mul(diff, bino(mul(n, 2), n));
  return 0;
}
