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;
inline int read() {
  int x = 0, f = 1;
  char ch = 0;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 998244353;
const int maxn = 2e5 + 100;
int T, n, m;
long long p, q;
long long a[maxn << 1];
long long f[maxn << 1], inv[maxn << 1];
inline long long C(int n, int m) {
  if (n < 0 || m < 0 || m > n)
    return 0;
  else
    return f[n] * inv[m] % mod * inv[n - m] % mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  f[0] = 1;
  for (int i = 1; i < maxn * 2; i++) f[i] = f[i - 1] * i % mod;
  inv[0] = inv[1] = 1;
  for (int i = 2; i < maxn * 2; i++)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  for (int i = 2; i < maxn * 2; i++) inv[i] = inv[i - 1] * inv[i] % mod;
  cin >> n;
  for (int i = 1; i <= n * 2; i++) cin >> a[i];
  sort(a + 1, a + 1 + n * 2);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans - a[i]) % mod;
  }
  for (int i = n + 1; i <= n * 2; i++) {
    ans = (ans + a[i]) % mod;
  }
  ans = ans * C(n * 2, n) % mod;
  cout << ans << endl;
  return 0;
}
