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;
bool comp1(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.first < p2.first || (p1.first == p2.first && p1.second < p2.second);
}
bool comp2(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.second < p2.second ||
         (p1.second == p2.second && p1.first < p2.first);
}
int main() {
  int n;
  scanf("%d", &n);
  pair<int, int> t[n];
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    t[i] = make_pair(x, y);
  }
  long long fact[n + 1];
  fact[0] = 1;
  for (int i = 0; i < n; i++) fact[i + 1] = (fact[i] * (i + 1)) % MOD;
  sort(t, t + n, comp1);
  bool sorted = true;
  for (int i = 0; i < n - 1; i++) {
    if (t[i].second > t[i + 1].second) sorted = false;
  }
  long long res = fact[n];
  long long inter = 0;
  if (sorted) inter = 1;
  long long sup = 1;
  int id = 0;
  while (id < n) {
    int id_deb = id;
    while (id < n - 1 && t[id + 1].first == t[id].first) {
      int idBis = id;
      while (id < n - 1 && t[id + 1].first == t[id].first &&
             t[id + 1].second == t[id].second)
        id++;
      inter *= fact[id + 1 - idBis];
      inter %= MOD;
      if (id < n - 1 && t[id + 1].first == t[id].first) id++;
    }
    sup *= fact[id + 1 - id_deb];
    sup %= MOD;
    id++;
  }
  res -= sup;
  res += inter;
  res %= MOD;
  sort(t, t + n, comp2);
  sup = 1;
  id = 0;
  while (id < n) {
    int id_deb = id;
    while (id < n - 1 && t[id + 1].second == t[id].second) {
      id++;
    }
    sup *= fact[id + 1 - id_deb];
    sup %= MOD;
    id++;
  }
  res -= sup;
  res %= MOD;
  if (res < 0) res += MOD;
  cout << res << endl;
  return 0;
}
