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 "trapv"
using namespace std;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.push_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
template <typename T_vector>
void output_vector(const T_vector& v, bool add_one = false, int start = -1,
                   int end = -1) {
  if (start < 0) start = 0;
  if (end < 0) end = int(v.sizes());
  for (int i = start; i < end; i++)
    cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
}
const long long N = 3e5 + 5;
unsigned long long fac[N + 1];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t clk = clock();
  fac[0] = 1;
  for (int i = 1; i <= N; i++) {
    fac[i] = (fac[i - 1] * i) % 998244353;
  }
  int t = 1;
  for (int tc = 0; tc < t; tc++) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> vec;
    map<long long, long long> ma, mb;
    map<pair<long long, long long>, long long> mvec;
    for (int i = 0; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      vec.push_back({a, b});
      ma[a]++;
      mb[b]++;
      mvec[{a, b}]++;
    }
    sort(vec.begin(), vec.end());
    bool ok = false;
    for (int i = 1; i < n; i++) {
      if (vec[i].first >= vec[i - 1].first &&
          vec[i].second >= vec[i - 1].second)
        continue;
      ok = true;
      break;
    }
    if (ok) {
      long long ans = fac[n];
      long long res = 1;
      for (auto x : ma) {
        res *= fac[x.second];
        res %= 998244353;
      }
      ans -= res;
      ans += 998244353;
      ans %= 998244353;
      res = 1;
      for (auto x : mb) {
        res *= fac[x.second];
        res %= 998244353;
      }
      ans -= res;
      ans += 998244353;
      ans %= 998244353;
      res = 1;
      cout << ans << endl;
    } else {
      long long ans = fac[n];
      long long res = 1;
      for (auto x : ma) {
        res *= fac[x.second];
        res %= 998244353;
      }
      ans -= res;
      ans += 998244353;
      ans %= 998244353;
      res = 1;
      for (auto x : mb) {
        res *= fac[x.second];
        res %= 998244353;
      }
      ans -= res;
      ans += 998244353;
      ans %= 998244353;
      res = 1;
      for (auto x : mvec) {
        res *= fac[x.second];
        res %= 998244353;
      }
      ans += res;
      ans += 998244353;
      ans %= 998244353;
      res = 1;
      cout << ans << endl;
    }
  }
  cerr << '\n'
       << "Time (in s): " << double(clock() - clk) * 1.0 / CLOCKS_PER_SEC
       << '\n';
}
