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("Ofast,unroll-loops,no-stack-protector")
using namespace std;
using ld = long double;
using lli = long long;
using ii = pair<int, int>;
using vi = vector<int>;
template <int p>
struct Modular {
  Modular operator+(const Modular &m) const { return Modular(v + m.v); }
  Modular operator-(const Modular &m) const { return Modular(v - m.v + p); }
  Modular operator*(const Modular &m) const { return Modular(1LL * v * m.v); }
  Modular inv() const { return Modular(minv(v, p)); }
  Modular operator-() const { return Modular(-v); }
  Modular &operator+=(const Modular &m) { return *this = *this + m; }
  Modular &operator-=(const Modular &m) { return *this = *this - m; }
  Modular &operator*=(const Modular &m) { return *this = *this * m; }
  friend ostream &operator<<(ostream &os, const Modular &m) {
    return os << m.v;
  }
  friend istream &operator>>(istream &is, Modular &m) { return is >> m.v; }
  int v;
  Modular(lli x = 0) : v((x %= p) < 0 ? x + p : x) {}

 private:
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - lli(minv(m, a)) * lli(m) / a);
  }
};
using Mint = Modular<998244353>;
Mint fac(int i) {
  static vector<Mint> dp(1, 1);
  while (dp.size() <= i) dp.push_back(dp.back() * dp.size());
  return dp[i];
}
const int N = 3e5 + 5;
ii second[N];
int n;
int main() {
  cin.tie(0)->sync_with_stdio(0), cout.tie(0);
  while (cin >> n) {
    for (auto i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
         i += 1 - 2 * ((0) > (n))) {
      cin >> second[i].first >> second[i].second;
    }
    Mint ans = fac(n);
    for (auto it = (0) - ((0) > (2)); it != (2) - ((0) > (2));
         it += 1 - 2 * ((0) > (2))) {
      map<int, int> cnt;
      for (auto i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
           i += 1 - 2 * ((0) > (n))) {
        cnt[second[i].first]++;
        swap(second[i].first, second[i].second);
      }
      Mint bad = 1;
      ;
      for (auto [x, k] : cnt) bad *= fac(k);
      ans -= bad;
    }
    sort(second, second + n);
    Mint common = 1;
    for (auto i = (1) - ((1) > (n)); i != (n) - ((1) > (n));
         i += 1 - 2 * ((1) > (n)))
      if (second[i].second < second[i - 1].second) {
        common = 0;
      }
    {
      map<ii, int> cnt;
      for (auto i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
           i += 1 - 2 * ((0) > (n)))
        cnt[second[i]]++;
      ;
      for (auto [x, k] : cnt) common *= fac(k);
      ans += common;
    }
    cout << ans << '\n';
  }
  return 0;
}
