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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 998244353;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int inv(long long int a) { return power(a, mod - 2); }
long long int add(long long int a, long long int b) { return (a + b) % mod; }
long long int sub(long long int a, long long int b) {
  return (a - b + mod) % mod;
}
long long int mul(long long int a, long long int b) { return (a * b) % mod; }
long long int divi(long long int a, long long int b) {
  return (a * inv(b)) % mod;
}
long long int fact[1000001];
long long int ifact[1000001];
void pre() {
  fact[0] = 1;
  ifact[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    fact[i] = mul(fact[i - 1], i);
    ifact[i] = power(fact[i], mod - 2);
  }
}
long long int ncr(int n, int r) {
  if (r > n) {
    return 0;
  }
  return divi(fact[n], mul(fact[r], fact[n - r]));
}
void solve() {
  long long int n, d;
  cin >> n;
  vector<long long int> a;
  for (int i = 0; i < 2 * n; i++) {
    cin >> d;
    a.push_back(d);
  }
  sort(a.begin(), a.end());
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans = sub(ans, mul(ncr(2 * n, n), a[i]));
    ans %= mod;
  }
  for (int i = n; i < 2 * n; i++) {
    ans = add(ans, mul(ncr(2 * n, n), a[i]));
    ans %= mod;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}
