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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, const pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, const TT<T...>& c) {
  out << "{ ";
  for (const auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const long long LIM = 1e5 + 5, MOD = 998244353;
const long double EPS = 1e-9;
const long long INF = 1e5;
vector<vector<int>> g;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> fac(n + 5);
  fac[1] = 1;
  fac[0] = 1;
  for (int i = 2; i < fac.size(); ++i) {
    fac[i] = (fac[i - 1] * (long long)i) % MOD;
  }
  vector<pair<long long, long long>> sx;
  vector<pair<long long, long long>> sy;
  for (int i = 0; i < n; ++i) {
    long long x, y;
    cin >> x >> y;
    sx.push_back({x, y});
    sy.push_back({y, x});
  }
  sort(sx.begin(), sx.end());
  sort(sy.begin(), sy.end());
  long long ans = 0;
  int k = 0;
  int t = 0;
  int g = 0;
  long long c1 = 1, c2 = 1, c3 = 1;
  bool f = false;
  for (int i = 1; i < n; ++i) {
    if (sx[i].first == sx[i - 1].first) {
      k += 1;
    } else {
      c1 = (c1 * fac[k + 1]) % MOD;
      k = 0;
    }
    if (sy[i].first == sy[i - 1].first) {
      t += 1;
    } else {
      c2 = (c2 * fac[t + 1]) % MOD;
      t = 0;
    }
    if (sx[i - 1].second > sx[i].second) {
      f = true;
    }
    if (!f) {
      if (sx[i].first == sx[i - 1].first && sx[i - 1].second == sx[i].second) {
        g += 1;
      } else {
        c3 = (c3 * fac[g + 1]) % MOD;
        g = 0;
      }
    }
  }
  c1 = (c1 * fac[k + 1]) % MOD;
  c2 = (c2 * fac[t + 1]) % MOD;
  c3 = (c3 * fac[g + 1]) % MOD;
  ans = (c1 + c2) % MOD;
  if (!f) {
    ans = (ans - c3 + MOD) % MOD;
  }
  cout << (fac[n] - ans + MOD) % MOD << '\n';
  return 0;
}
