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 int m = 998244353;
long long int fact(long long int n) {
  long long int fac = 1;
  for (long long int i = 1; i <= n; i++) {
    fac = ((fac % m) * (i % m)) % m;
  }
  return fac;
}
bool compare(pair<long long int, long long int> a,
             pair<long long int, long long int> b) {
  if (a.first != b.first)
    return a.first < b.first;
  else
    return a.second < b.second;
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  long long int b[n + 1];
  pair<long long int, long long int> pairs[n + 1];
  map<pair<long long int, long long int>, long long int> m3;
  map<long long int, long long int> m1;
  map<long long int, long long int> m2;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    m1[a[i]]++;
    m2[b[i]]++;
    m3[(make_pair(a[i], b[i]))]++;
    pairs[i] = make_pair(a[i], b[i]);
  }
  long long int first = 1;
  long long int second = 1;
  vector<long long int> cs;
  sort(pairs + 1, pairs + n + 1, compare);
  long long int f = 1;
  for (long long int i = 1; i <= n - 1; i++) {
    if (pairs[i + 1].second < pairs[i].second) {
      f = 0;
      break;
    }
  }
  long long int both = 1;
  if (!f) both = 0;
  for (auto x : m1) {
    first = (first * fact(x.second)) % m;
  }
  for (auto x : m2) {
    second = (second * fact(x.second)) % m;
  }
  for (auto x : m3) {
    both = (both * fact(x.second)) % m;
  }
  long long int sub = (first % m + second % m - both % m + m) % m;
  cout << (fact(n) % m - sub % m + m) % m;
}
