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;
pair<int, int> input[300003];
long long factor[300003];
int num1[300003];
int num2[300003];
int suc = 1;
int compare(pair<int, int> a, pair<int, int> b) {
  if (a.first >= b.first && a.second >= b.second)
    return 0;
  else if (a.first <= b.first && a.second <= b.second)
    return 1;
  else {
    suc = 0;
    return 0;
  }
}
int main() {
  factor[0] = factor[1] = 1;
  for (int i = 2; i < 300003; i++) factor[i] = (factor[i - 1] * i) % 998244353;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> input[i].first >> input[i].second;
    num1[input[i].first]++;
    num2[input[i].second]++;
  }
  long long sum1 = 1;
  long long sum2 = 1;
  for (int i = 1; i <= n; i++) {
    if (num1[i] != 0) sum1 = (sum1 * factor[num1[i]]) % 998244353;
    if (num2[i] != 0) sum2 = (sum2 * factor[num2[i]]) % 998244353;
  }
  sort(input, input + n, compare);
  long long total_sum = 1;
  if (suc == 0)
    total_sum = 0;
  else {
    int last1 = input[0].first;
    int last2 = input[0].second;
    int c = 1;
    for (int i = 1; i < n; i++) {
      if (!(input[i].first >= last1 && input[i].second >= last2)) {
        total_sum = 0;
        break;
      }
      if (input[i].first == last1 && input[i].second == last2)
        c++;
      else {
        total_sum = (total_sum * factor[c]) % 998244353;
        c = 1;
      }
      last1 = input[i].first;
      last2 = input[i].second;
    }
    total_sum = (total_sum * factor[c]) % 998244353;
  }
  cout << ((factor[n] - sum1 - sum2 + total_sum) % 998244353 + 998244353) %
              998244353
       << endl;
  return 0;
}
