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;
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "{" << p.first << "," << p.second << "}";
  return os;
}
const long long N = 3e5 + 5;
const long long oo = 998244353;
long long extended_euclidean(long long a, long long b, long long &x,
                             long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = extended_euclidean(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mod_inverse(long long a, long long m) {
  long long x, y;
  long long g = extended_euclidean(a, m, x, y);
  if (g != 1) {
    cout << "No solution!";
    return -1;
  } else {
    x = (x % m + m) % m;
    return x;
  }
}
long long fact[N];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = i * fact[i - 1] % oo;
  }
  long long n;
  cin >> n;
  map<long long, long long> x;
  map<long long, long long> y;
  map<pair<long long, long long>, long long> xy;
  vector<pair<long long, long long> > xxy;
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    x[a]++;
    y[b]++;
    xy[{a, b}]++;
    xxy.push_back({a, b});
  }
  sort(xxy.begin(), xxy.end());
  long long p1 = 1;
  for (auto i : x) {
    p1 = p1 * (fact[i.second]) % oo;
  }
  long long ans = fact[n];
  ans -= p1;
  ans %= oo;
  p1 = 1;
  for (auto i : y) {
    p1 = p1 * (fact[i.second]) % oo;
  }
  ans -= p1;
  ans %= oo;
  p1 = 1;
  long long d = 1;
  for (long long i = 1; i < n; i++) {
    if (xxy[i].first < xxy[i - 1].first or xxy[i].second < xxy[i - 1].second)
      d = 0;
  }
  for (auto i : xy) {
    p1 = p1 * (fact[i.second]) % oo;
  }
  if (d) ans += p1;
  ans %= oo;
  ans += oo;
  ans %= oo;
  cout << ans << "\n";
}
