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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int first = 0;
  cout << '{';
  for (auto &i : x) cout << (first++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
template <class T>
void remdup(vector<T> &v) {
  sort((v).begin(), (v).end());
  v.erase(unique((v).begin(), (v).end()), end(v));
}
template <typename T>
struct cmp {
  bool operator()(const T &p1, const T &p2) {}
};
const long long N = 3e5 + 100;
const long long mod = 998244353;
const long long X = 1e5 + 100;
const long long Y = 1e9 + 100;
long long power(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return result;
}
long long pinverse(long long A, long long M) { return power(A, M - 2); }
long long fact[N], invfact[N];
void precompute() {
  fact[0] = fact[1] = 1;
  for (long long i = (2); i < (N); ++i) fact[i] = (fact[i - 1] * i) % mod;
  invfact[N - 1] = pinverse(fact[N - 1], mod);
  for (long long i = (N - 2); i >= (0); --i)
    invfact[i] = (invfact[i + 1] * (i + 1)) % mod;
}
long long nCr(long long x, long long y) {
  if (y > x) return 0;
  long long num = fact[x];
  num = (num * invfact[y]) % mod;
  num = (num * invfact[x - y]) % mod;
  return num;
}
int main() {
  long long t = 1;
  precompute();
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(2 * n);
    for (long long i = (0); i < (2 * n); ++i) cin >> a[i];
    sort((a).begin(), (a).end());
    long long sum = 0;
    for (long long i = (0); i < (2 * n); ++i) sum += ((i < n) ? -a[i] : a[i]);
    sum %= mod;
    sum *= nCr(2 * n, n);
    sum %= mod;
    cout << sum << "\n";
  }
  return 0;
}
