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>
const long long mod = 998244353;
int a[300005];
inline int read() {
  register int x = 0, f = 1;
  register char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
inline long long pow(long long x, long long p) {
  long long res = 1;
  for (; p; p >>= 1) {
    if (p & 1) res = res * x % mod;
    x = x * x % mod;
  }
  return res;
}
int main() {
  int n = read();
  long long ans = 0, tmp1 = 1, tmp2 = 1;
  for (register int i = 1; i <= n; ++i) tmp1 = tmp1 * i % mod;
  tmp2 = tmp1;
  for (register int i = n + 1; i <= 2 * n; ++i) tmp2 = tmp2 * i % mod;
  tmp1 = tmp1 * tmp1 % mod;
  for (register int i = 1; i <= 2 * n; ++i) a[i] = read();
  std::sort(a + 1, a + 1 + 2 * n);
  for (register int i = 1; i <= n; ++i) ans += a[n + i] - a[i];
  ans %= mod;
  printf("%lld\n", ans * tmp2 % mod * pow(tmp1, mod - 2) % mod);
  return 0;
}
