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;
class dop {
 public:
  long long int n, p, q, r, f, g;
  vector<long long int> a, b;
  vector<pair<long long int, long long int>> c;
  vector<long long int> d, e;
  void get();
};
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  if (a.first < b.first) {
    return true;
  } else if (a.first == b.first) {
    if (a.second < b.second) {
      return true;
    }
  }
  return false;
}
bool compe(pair<long long int, long long int> a,
           pair<long long int, long long int> b) {
  if (a.second < b.second) {
    return true;
  } else if (a.second == b.second) {
    if (a.first < b.first) {
      return true;
    }
  }
  return false;
}
void dop::get() {
  cin >> n;
  vector<long long int> fact(n + 1, 0);
  fact[0] = 1;
  p = 1;
  for (long long int i = 0; i < n; i++) {
    p = ((p % 998244353) * ((i + 1) % 998244353)) % 998244353;
    fact[i + 1] = p % 998244353;
  }
  for (long long int i = 0; i < n; i++) {
    cin >> p >> q;
    a.push_back(p);
    b.push_back(q);
    c.push_back(make_pair(p, q));
  }
  sort(c.begin(), c.end(), comp);
  for (long long int i = 0; i < n; i++) {
    d.push_back(c[i].first);
    e.push_back(c[i].second);
  }
  p = 1;
  sort(a.begin(), a.end());
  for (long long int i = 0; i < n; i++) {
    auto ptr = upper_bound(a.begin(), a.end(), a[i]);
    p = ((p % 998244353) * (fact[ptr - a.begin() - i] % 998244353)) % 998244353;
    i = ptr - a.begin() - 1;
  }
  q = 1;
  sort(b.begin(), b.end());
  for (long long int i = 0; i < n; i++) {
    auto ptr = upper_bound(b.begin(), b.end(), b[i]);
    q = ((q % 998244353) * (fact[ptr - b.begin() - i] % 998244353)) % 998244353;
    i = ptr - b.begin() - 1;
  }
  p = (p + q) % 998244353;
  p = (p % 998244353);
  if (p < 0) {
    p += 998244353;
  }
  q = 1;
  if (d == a && b == e) {
    for (long long int i = 0; i < n; i++) {
      auto ptr = upper_bound(d.begin(), d.end(), d[i]);
      auto itr = upper_bound(e.begin(), e.end(), e[i]);
      g = ptr - d.begin() - i;
      f = itr - e.begin() - i;
      if (g < f) {
        q = ((q % 998244353) * (fact[g] % 998244353)) % 998244353;
        i = ptr - d.begin() - 1;
      } else {
        q = ((q % 998244353) * (fact[f] % 998244353)) % 998244353;
        i = itr - e.begin() - 1;
      }
    }
    q = (q + fact[n]) % 998244353;
    q = (q - p);
    if (q < 0) {
      q += 998244353;
    }
    q = q % 998244353;
    cout << q << '\n';
    return;
  }
  q = fact[n] - p;
  q = q % 998244353;
  if (q < 0) {
    q += 998244353;
  }
  cout << q << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  dop obj[t];
  for (long long int i = 0; i < t; i++) {
    obj[i].get();
  }
  return 0;
}
