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 MOD = 998244353;
long long modpow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= MOD;
    }
    a *= a;
    a %= MOD;
    b /= 2;
  }
  return ans;
}
long long modinv(long long a) { return modpow(a, MOD - 2); }
vector<long long> mf = {1};
vector<long long> mfi = {1};
long long modfact(int n) {
  if (mf.size() > n) {
    return mf[n];
  } else {
    for (int i = mf.size(); i <= n; i++) {
      long long next = mf.back() * i % MOD;
      mf.push_back(next);
      mfi.push_back(modinv(next));
    }
    return mf[n];
  }
}
long long modfactinv(int n) {
  if (mfi.size() > n) {
    return mfi[n];
  } else {
    return modinv(modfact(n));
  }
}
long long modbinom(int n, int k) {
  if (n < 0 || k < 0 || k > n) {
    return 0;
  } else {
    return modfact(n) * modfactinv(k) % MOD * modfactinv(n - k) % MOD;
  }
}
int main() {
  int n;
  cin >> n;
  vector<long long> a(n * 2);
  for (int i = 0; i < n * 2; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  for (int i = n; i < n * 2; i++) {
    ans += a[i];
  }
  for (int i = 0; i < n; i++) {
    ans -= a[i];
  }
  ans *= 2;
  ans %= MOD;
  ans *= modbinom(n * 2 - 1, n - 1);
  ans %= MOD;
  cout << ans << endl;
}
