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 N = 305000;
const long long mod = 998244353;
long long n;
long long num[N];
long long ans;
inline long long read() {
  long long re = 0, k = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') k = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    re = re * 10 + ch - '0';
    ch = getchar();
  }
  return re * k;
}
inline long long ksm(long long x, long long y) {
  long long sum = 1;
  while (y) {
    if (y & 1) sum = (sum * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return sum;
}
inline long long inv(long long x) { return ksm(x, mod - 2); }
inline long long f(long long x) {
  long long sum = 1;
  for (register long long i = 1; i <= x; i++) sum = (sum * i) % mod;
  return sum;
}
inline long long calc(long long x, long long y) {
  return f(x) * inv(f(y)) % mod * inv(f(x - y)) % mod;
}
signed main() {
  n = read();
  for (register long long i = 1; i <= n + n; i++) num[i] = read();
  sort(num + 1, num + 1 + n + n);
  for (register long long i = 1; i <= n; i++) ans -= num[i];
  for (register long long i = n + 1; i <= 2 * n; i++) ans += num[i];
  printf("%lld", ans % mod * calc(n + n, n) % mod);
  return 0;
}
