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;
long long mod = 998244353L;
int n;
vector<pair<long long, long long> > arr;
void swap_pairs() {
  for (int i = 0; i < n; i++) swap(arr[i].first, arr[i].second);
}
long long fact[1000000];
void calcule_factorial() {
  fact[0] = 1;
  for (long long i = 1; i < 1000000; i++) fact[i] = (i * fact[i - 1]) % mod;
}
long long count_bad_by_first_element() {
  sort(arr.begin(), arr.end());
  map<long long, long long> m;
  for (int i = 0; i < n; i++) {
    m[arr[i].first]++;
  }
  long long ret = 1;
  map<long long, long long>::iterator it = m.begin();
  while (it != m.end()) {
    pair<long long, long long> kv = *(it);
    ret = (ret * fact[kv.second]) % mod;
    it++;
  }
  return ret;
}
bool is_sorted_by_first_and_second_element() {
  for (int i = 1; i < n; i++) {
    if (arr[i - 1].second > arr[i].second) return false;
  }
  return true;
}
long long count_bad_by_first_and_second_element() {
  sort(arr.begin(), arr.end());
  if (!is_sorted_by_first_and_second_element()) {
    return 0;
  }
  map<pair<long long, long long>, long long> m;
  for (int i = 0; i < n; i++) {
    m[arr[i]]++;
  }
  long long ret = 1;
  map<pair<long long, long long>, long long>::iterator it = m.begin();
  while (it != m.end()) {
    pair<pair<long long, long long>, long long> kv = *(it);
    ret = (ret * fact[kv.second]) % mod;
    it++;
  }
  return ret;
}
void solve_test_case() {
  calcule_factorial();
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    arr.push_back(make_pair(a, b));
  }
  long long bad_by_first_element = count_bad_by_first_element();
  swap_pairs();
  long long bad_by_second_element = count_bad_by_first_element();
  swap_pairs();
  long long bad_by_first_and_second_element =
      count_bad_by_first_and_second_element();
  long long ans = (fact[n] - bad_by_first_element + mod) % mod;
  ans = (ans - bad_by_second_element + mod) % mod;
  ans = (ans + bad_by_first_and_second_element) % mod;
  cout << ans << '\n';
}
void solve_test_cases() { solve_test_case(); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve_test_cases();
  return 0;
}
