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 maxn = 150005;
const long long inf = 0x3f3f3f3f;
const long long p = 998244353;
long long N, sum;
long long a[maxn * 2];
long long qpow(long long u, long long v) {
  long long ret = 1;
  while (v) {
    if (v & 1) ret = ret * u % p;
    u = u * u % p;
    v >>= 1;
  }
  return ret;
}
long long C(long long n, long long m) {
  if (m > n) return 0;
  long long ans = 1;
  for (int i = 1; i <= m; i++) {
    long long u = (n + i - m) % p;
    long long v = i % p;
    ans = ans * (u * qpow(v, p - 2) % p) % p;
  }
  return ans;
}
long long Lucas(long long n, long long m) {
  if (m == 0) return 1;
  return C(n % p, m % p) * Lucas(n / p, m / p) % p;
}
int main() {
  scanf("%lld", &N);
  for (int i = 1; i <= N * 2; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + N * 2 + 1);
  int l = 1, r = N * 2;
  while (l < r) {
    sum = (sum + ((a[r] - a[l]) % p)) % p;
    ++l;
    --r;
  }
  printf("%lld\n", sum * Lucas(N * 2, N) % p);
  return 0;
}
