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;
bool mc(pair<long long int, long long int> p1,
        pair<long long int, long long int> p2) {
  if (p1.first == p2.first) return p1.second < p2.second;
  return p1.first < p2.first;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long int> fac(n + 1, 0);
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % 998244353;
  vector<pair<long long int, long long int> > a(n);
  for (auto &it : a) cin >> it.first >> it.second;
  long long ans = fac[n];
  vector<long long int> tf(n + 1, 0), ts(n + 1, 0);
  for (auto it : a) tf[it.first]++, ts[it.second]++;
  long long ff = 1, fs = 1, ffs = 1;
  for (auto it : tf) ff = (ff * fac[it]) % 998244353;
  for (auto it : ts) fs = (fs * fac[it]) % 998244353;
  ans = (ans - ff + 998244353) % 998244353;
  ans = (ans - fs + 998244353) % 998244353;
  sort(a.begin(), a.end(), mc);
  long long i = 0;
  bool first = true;
  for (long long j = 0; j < n - 1; j++)
    if (a[j].second > a[j + 1].second) first = false;
  if (!first) i = n + 1, ffs = 0;
  vector<long long int> temts(n + 1, 0), vis(n + 1, 0);
  while (i < n) {
    vector<long long int> second;
    for (long long j = i; j < i + tf[a[i].first]; j++) {
      if (!vis[a[j].second]) second.push_back(a[j].second);
      vis[a[j].second] = 1, temts[a[j].second]++;
    }
    for (auto it : second) {
      ffs = (ffs * fac[temts[it]]) % 998244353;
      vis[it] = 0, temts[it] = 0;
    }
    i += tf[a[i].first];
  }
  ans = (ans + ffs) % 998244353;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  long long t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  cin.get();
  return 0;
}
