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 long long MX = 3e5 + 5;
const long long M = 998244353;
long long n, p, q, d, x, a[MX], f[MX], inv[MX];
long long pw(long long x, long long y) {
  long long r = 1;
  while (y) {
    if (y & 1) r = (r * x) % M;
    y >>= 1;
    x = (x * x) % M;
  }
  if (r < 0) r += M;
  return r;
}
long long ncr(long long n, long long r) {
  long long x = f[n];
  x *= inv[r], x %= M;
  x *= inv[n - r], x %= M;
  if (x < 0) x += M;
  return x;
}
int main() {
  f[0] = 1;
  for (long long i = 1; i < MX; i++) f[i] = (f[i - 1] * i) % M;
  inv[MX - 1] = pw(f[MX - 1], M - 2);
  for (long long i = MX - 2; i; i--) inv[i] = (inv[i + 1] * (i + 1)) % M;
  cin >> n;
  for (long long i = 0; i < 2 * n; i++) cin >> a[i];
  sort(a, a + 2 * n);
  for (long long i = 0; i < 2 * n; i++) {
    if (i < n)
      q += a[i];
    else
      p += a[i];
  }
  d = (p - q) % M;
  x = ncr(2 * n, n);
  x *= d;
  x %= M;
  if (x < 0) x += M;
  cout << x;
  return 0;
}
