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 N = 300009;
const long long mod = 998244353;
const long long mod2 = 1e6;
long long mul(long long x, long long y) {
  x %= mod, y %= mod;
  x = (x + mod) % mod, y = (y + mod) % mod;
  long long p = y / mod2;
  long long ans = (y % mod2 * x % mod + (p * x) % mod * mod2 % mod) % mod;
  ans = (ans + mod) % mod;
  return ans;
}
long long n, a[N];
long long fac[N], inv[N];
long long C(int x, int y) { return mul(mul(fac[y], inv[x]), inv[y - x]); }
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= 2 * n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 2 * n + 1);
  for (int i = 1; i <= 2 * n; i++) a[i] %= mod;
  fac[0] = 1;
  for (int i = 1; i <= 2 * n; i++) fac[i] = mul(fac[i - 1], i);
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= 2 * n; i++) inv[i] = mul(inv[mod % i], (mod - mod / i));
  for (int i = 1; i <= 2 * n; i++) inv[i] = mul(inv[i], inv[i - 1]);
  long long ans = 0, val = C(n - 1, 2 * n - 1);
  for (int i = 1; i < n; i++) {
    ans = (ans + mul(-a[i], val)) % mod;
  }
  ans = (ans - mul(a[n], val));
  ans = (ans % mod + mod) % mod;
  for (int i = n + 1; i <= 2 * n; i++) {
    ans = (ans + mul(a[i], val)) % mod;
  }
  printf("%lld\n", ans * 2ll % mod);
  return 0;
}
