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 factmodMAX(int ma) {
  long long rez = 1;
  for (long long i = 2; i <= ma; i++) {
    rez *= i;
    if (rez > 998244353) rez %= 998244353;
  }
  return rez;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  long long ans, badl, badr, bad2;
  ans = factmodMAX(n);
  vector<int> ri;
  vector<pair<int, int> > leri;
  int bufl, bufr;
  for (int i = 0; i < n; i++) {
    cin >> bufl >> bufr;
    ri.push_back(bufr);
    leri.push_back(pair<int, int>(bufl, bufr));
  }
  sort(leri.begin(), leri.end());
  leri.push_back(pair<int, int>(1000000, 1000000));
  int ind1 = 0;
  badl = 1;
  for (int i = 0; i <= n; i++) {
    if (leri[i].first == leri[ind1].first) continue;
    badl *= factmodMAX(i - ind1);
    badl %= 998244353;
    ind1 = i;
  }
  bool bad2e = true;
  for (int i = 1; i <= n; i++) {
    if (leri[i].second >= leri[i - 1].second) continue;
    bad2e = false;
    break;
  }
  if (bad2e) {
    bad2 = 1;
    ind1 = 0;
    for (int i = 0; i <= n; i++) {
      if (leri[i].first == leri[ind1].first &&
          leri[i].second == leri[ind1].second)
        continue;
      bad2 *= factmodMAX(i - ind1);
      bad2 %= 998244353;
      ind1 = i;
    }
  } else {
    bad2 = 0;
  }
  sort(ri.begin(), ri.end());
  ri.push_back(1000000);
  ind1 = 0;
  badr = 1;
  for (int i = 0; i <= n; i++) {
    if (ri[i] == ri[ind1]) continue;
    badr *= factmodMAX(i - ind1);
    badr %= 998244353;
    ind1 = i;
  }
  ans = ans - badl - badr + bad2 + (long long)(998244353 * 2);
  ans %= 998244353;
  cout << ans << endl;
  return 0;
}
