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 = 300005, mo = 998244353;
int i, j, n, ans;
long long ch, en;
int a[N], jc[N], ny[N];
int read() {
  int x = 0;
  char s = getchar();
  while ((s < '0') || (s > '9')) s = getchar();
  while ((s >= '0') && (s <= '9')) x = x * 10 + (s - '0'), s = getchar();
  return x;
}
int ksm(int x, int y) {
  long long l = x, r = 1;
  int z = y;
  while (z) {
    if (z % 2 == 1) r = r * l % mo;
    l = l * l % mo;
    z /= 2;
  }
  return r;
}
void kp(int x, int y) {
  int l = x, r = y, mid = a[(l + r) / 2];
  while (l <= r) {
    while (a[l] < mid) l++;
    while (a[r] > mid) r--;
    if (l <= r) {
      a[0] = a[l];
      a[l] = a[r];
      a[r] = a[0];
      l++;
      r--;
    }
  }
  if (l < y) kp(l, y);
  if (x < r) kp(x, r);
}
int main() {
  n = read();
  jc[0] = 1;
  ny[0] = 1;
  ch = 1;
  for (i = 1; i <= 2 * n; i++) {
    ch = ch * i % mo;
    jc[i] = ch;
  }
  ch = ksm(ch, mo - 2);
  ny[2 * n] = ch;
  for (i = 2 * n - 1; i >= 1; i--) {
    ch = ch * (i + 1) % mo;
    ny[i] = ch;
  }
  en = jc[2 * n];
  en = en * ny[n] % mo * ny[n] % mo;
  for (i = 1; i <= 2 * n; i++) a[i] = read();
  kp(1, 2 * n);
  for (i = n + 1; i <= 2 * n; i++) ans = (ans + a[i] * en) % mo;
  for (i = 1; i <= n; i++) ans = (ans - a[i] * en % mo + mo) % mo;
  printf("%d\n", ans);
  fclose(stdin);
  fclose(stdout);
}
