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 = 300010, mod = 998244353;
int a[N];
int n;
int fac[N], inv[N];
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int del(int a, int b) { return a < b ? a - b + mod : a - b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
inline int qpow(int a, int b) {
  int s = 1;
  while (b) {
    if (b & 1) s = mul(s, a);
    a = mul(a, a);
    b >>= 1;
  }
  return s;
}
int main() {
  scanf("%d", &n);
  int m = 2 * n;
  fac[0] = fac[1] = 1;
  for (int i = 2; i <= m; ++i) fac[i] = mul(fac[i - 1], i);
  inv[m] = qpow(fac[m], mod - 2);
  for (int i = m - 1; i >= 1; --i) inv[i] = mul(inv[i + 1], i + 1);
  inv[0] = 1;
  for (int i = 1; i <= m; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + m);
  for (int i = 1; i <= m; ++i)
    if (a[i] >= mod) a[i] %= mod;
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = del(ans, a[i]);
  for (int i = n + 1; i <= m; ++i) ans = add(ans, a[i]);
  printf("%d\n", mul(ans, mul(fac[m], mul(inv[n], inv[n]))));
  return 0;
}
