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 long long mod = 998244353;
long long factorial(long long k) {
  long long ans = 1;
  for (long long i = 1; i <= k; i++) ans = ans * i % mod;
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  vector<long long> b(n);
  vector<long long> c(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    b[i] = a[i].first;
    c[i] = a[i].second;
  }
  sort(a.begin(), a.begin() + n);
  sort(b.begin(), b.begin() + n);
  sort(c.begin(), c.begin() + n);
  long long num1 = 1, num2 = 1, num3 = 1, counter = 1, flag = 1;
  for (int i = 0; i < n - 1; i++)
    if (b[i + 1] == b[i]) {
      counter++;
    } else if (counter > 1) {
      num1 = num1 * factorial(counter) % mod;
      counter = 1;
    }
  num1 = num1 * factorial(counter) % mod;
  counter = 1;
  for (int i = 0; i < n - 1; i++)
    if (c[i + 1] == c[i])
      counter++;
    else if (counter > 1) {
      num2 = num2 * factorial(counter) % mod;
      counter = 1;
    }
  num2 = num2 * factorial(counter) % mod;
  counter = 1;
  for (int i = 0; i < n - 1; i++)
    if (a[i + 1].second < a[i].second) flag = 0;
  if (flag) {
    for (int i = 0; i < n - 1; i++)
      if (a[i + 1].first == a[i].first && a[i + 1].second == a[i].second)
        counter++;
      else if (counter > 1) {
        num3 = num3 * factorial(counter) % mod;
        counter = 1;
      }
    num3 = num3 * factorial(counter) % mod;
  } else
    num3 = 0;
  cout << ((factorial(n) - num1 - num2 + num3) % mod + mod) % mod << endl;
  return 0;
}
