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 = 300005;
const long long mod = 998244353;
long long fac[maxn], invfac[maxn];
int a[maxn];
long long qp(long long x, long long p, long long MOD) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * x % MOD;
    x = x * x % MOD;
    p >>= 1;
  }
  return res;
}
void init() {
  fac[0] = 1;
  invfac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = fac[i - 1] * 1ll * i % mod;
  invfac[maxn - 1] = qp(fac[maxn - 1], mod - 2, mod);
  for (int i = maxn - 2; i >= 1; i--)
    invfac[i] = invfac[i + 1] * 1ll * (i + 1) % mod;
}
long long C(int n, int k) {
  return fac[n] * invfac[n - k] % mod * invfac[k] % mod;
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n + n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n * 2 + 1);
  long long ans = 0;
  for (int i = 1; i <= n * 2; i++) {
    if (i <= n) {
      ans = (ans - C(n * 2, n) * a[i] % mod) % mod;
    } else {
      ans = (ans + C(n * 2, n) * a[i] % mod + mod) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
