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;
template <typename T>
void read(T& x) {
  x = 0;
  int fl = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fl = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  x *= fl;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const int N = 300005;
const long long MOD = 998244353;
int n;
long long a[N];
inline long long fpow(long long bs, long long idx) {
  long long res = 1;
  bs %= MOD;
  while (idx) {
    if (idx & 1) (res *= bs) %= MOD;
    (bs *= bs) %= MOD;
    idx >>= 1;
  }
  return res;
}
int main() {
  read(n);
  for (int i = 1; i <= n * 2; i++) read(a[i]);
  sort(a + 1, a + n * 2 + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) (ans += a[i + n] - a[i]) %= MOD;
  for (int i = 1; i <= n; i++) {
    (ans *= (i + n)) %= MOD;
    (ans *= fpow(i, MOD - 2)) %= MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
