You are given a sequence of n pairs of integers: (a_1, b_1), (a_2, b_2), ... , (a_n, b_n). This sequence is called bad if it is sorted in non-descending order by first elements or if it is sorted in non-descending order by second elements. Otherwise the sequence is good. There are examples of good and bad sequences:

  * s = [(1, 2), (3, 2), (3, 1)] is bad because the sequence of first elements is sorted: [1, 3, 3]; 
  * s = [(1, 2), (3, 2), (1, 2)] is bad because the sequence of second elements is sorted: [2, 2, 2]; 
  * s = [(1, 1), (2, 2), (3, 3)] is bad because both sequences (the sequence of first elements and the sequence of second elements) are sorted; 
  * s = [(1, 3), (3, 3), (2, 2)] is good because neither the sequence of first elements ([1, 3, 2]) nor the sequence of second elements ([3, 3, 2]) is sorted. 



Calculate the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence. 

A permutation p of size n is a sequence p_1, p_2, ... , p_n consisting of n distinct integers from 1 to n (1 ≤ p_i ≤ n). If you apply permutation p_1, p_2, ... , p_n to the sequence s_1, s_2, ... , s_n you get the sequence s_{p_1}, s_{p_2}, ... , s_{p_n}. For example, if s = [(1, 2), (1, 3), (2, 3)] and p = [2, 3, 1] then s turns into [(1, 3), (2, 3), (1, 2)].

Input

The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5).

The next n lines contains description of sequence s. The i-th line contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — the first and second elements of i-th pair in the sequence.

The sequence s may contain equal elements.

Output

Print the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence. Print the answer modulo 998244353 (a prime number).

Examples

Input


3
1 1
2 2
3 1


Output


3


Input


4
2 3
2 2
2 1
2 4


Output


0


Input


3
1 1
1 1
2 3


Output


4

Note

In first test case there are six permutations of size 3: 

  1. if p = [1, 2, 3], then s = [(1, 1), (2, 2), (3, 1)] — bad sequence (sorted by first elements); 
  2. if p = [1, 3, 2], then s = [(1, 1), (3, 1), (2, 2)] — bad sequence (sorted by second elements); 
  3. if p = [2, 1, 3], then s = [(2, 2), (1, 1), (3, 1)] — good sequence; 
  4. if p = [2, 3, 1], then s = [(2, 2), (3, 1), (1, 1)] — good sequence; 
  5. if p = [3, 1, 2], then s = [(3, 1), (1, 1), (2, 2)] — bad sequence (sorted by second elements); 
  6. if p = [3, 2, 1], then s = [(3, 1), (2, 2), (1, 1)] — good sequence. 
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const long long mod = 998244353;
long long fun[N] = {1};
struct node {
  int xx, yy;
} a[N];
bool cmp(node i, node j) {
  if (i.xx == j.xx) return i.yy < j.yy;
  return i.xx < j.xx;
}
int x[N], y[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    fun[i] = fun[i - 1] * i % mod;
    scanf("%d %d", &a[i].xx, &a[i].yy);
    x[i] = a[i].xx;
    y[i] = a[i].yy;
  }
  sort(x + 1, x + n + 1);
  sort(y + 1, y + n + 1);
  sort(a + 1, a + n + 1, cmp);
  bool fg = false;
  long long sum = 1, sum1 = 1, sum2 = 1, num = 1, num1 = 1, num2 = 1;
  for (int i = 2; i <= n; i++) {
    if (x[i] == x[i - 1])
      num1++;
    else
      sum1 = sum1 * fun[num1] % mod, num1 = 1;
    if (y[i] == y[i - 1])
      num2++;
    else
      sum2 = sum2 * fun[num2] % mod, num2 = 1;
    if (a[i].xx == a[i - 1].xx && a[i].yy == a[i - 1].yy)
      num++;
    else
      sum = sum * fun[num] % mod, num = 1;
    if (a[i].yy < a[i - 1].yy) fg = true;
  }
  if (x[n] == x[n - 1]) sum1 = sum1 * fun[num1] % mod;
  if (y[n] == y[n - 1]) sum2 = sum2 * fun[num2] % mod;
  if (a[n].xx == a[n - 1].xx && a[n].yy == a[n - 1].yy)
    sum = sum * fun[num] % mod;
  if (fg) sum = 0;
  long long ans =
      (((fun[n] - sum1 + mod) % mod - sum2 + mod) % mod + sum) % mod;
  printf("%I64d\n", ans);
  return 0;
}
