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 <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "]\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto it : v) os << it.first << " : " << it.second << "\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  return os;
}
void go() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const long long N = 3e5 + 5;
const long long mod = 998244353;
long long f[N];
void pre() {
  f[0] = 1;
  for (long long i = 1; i < N; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
}
void solve() {
  pre();
  long long n;
  cin >> n;
  vector<pair<long long, long long>> p;
  vector<long long> a(n), b(n), fa(n + 5), fb(n + 5);
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    fa[a[i]]++;
    fb[b[i]]++;
    p.push_back({a[i], b[i]});
  }
  sort(p.begin(), p.end());
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  long long ff = 1, ss = 1;
  for (long long i = 0; i <= n; i++) {
    ff = (ff * f[fa[i]]) % mod;
    ss = (ss * f[fb[i]]) % mod;
  }
  long long tt = 0;
  for (long long i = 0; i < n; i++) {
    b[i] = p[i].second;
  }
  if (is_sorted(b.begin(), b.end())) {
    tt = 1;
    long long c = 1;
    for (long long i = 1; i < n; i++) {
      if (p[i].first == p[i - 1].first and p[i].second == p[i - 1].second) {
        c++;
      } else {
        tt = (tt * f[c]) % mod;
        c = 1;
      }
    }
    tt = (tt * f[c]) % mod;
  }
  cout << ((f[n] - ff - ss + tt + mod) % mod + mod) % mod;
}
int main() {
  go();
  solve();
}
