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 INF = 0x3f3f3f3f;
const long long INFLL = 2e18;
const int MOD = 998244353;
const int MAXN = 3e5 + 5;
int n, a[MAXN];
long long first[MAXN];
void fillF() {
  first[0] = first[1] = 1;
  for (int i = 2; i < 2 * n + 1; i++)
    first[i] = ((long long)i * first[i - 1]) % MOD;
}
long long sq(long long x) { return (x * x) % MOD; }
long long binpow(long long b, long long e) {
  if (e == 0) return 1LL;
  if (e == 1) return b;
  return (sq(binpow(b, e / 2LL)) * (e & 1LL ? b : 1LL)) % MOD;
}
long long C(long long n_, long long k) {
  long long inv = (first[k] * first[n_ - k]) % MOD;
  inv = binpow(inv, MOD - 2);
  return (first[n_] * inv) % MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 2 * n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + 2 * n);
  long long sum = 0;
  for (int i = 0; i < n; i++) sum = (sum - a[i] + MOD) % MOD;
  for (int i = n; i < 2 * n; i++) sum = (sum + a[i]) % MOD;
  fillF();
  printf("%lld\n", (C(2 * n, n) * sum) % MOD);
  return 0;
}
