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;
long long power(long long x, long long y, long long md = 998244353) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
long long min1() { return 1000000000000000LL; }
long long max1() { return -1000000000000000LL; }
template <typename... Args>
long long min1(long long a, Args... args) {
  return min(a, min1(args...));
}
template <typename... Args>
long long max1(long long a, Args... args) {
  return max(a, max1(args...));
}
struct node {
  std::vector<long long> v, w;
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<pair<long long, long long> > second(n);
  for (long long i = 0; i < n; i++) cin >> second[i].first >> second[i].second;
  long long first[300002];
  first[0] = 1;
  for (long long i = 1; i < 300002; i++)
    first[i] = (first[i - 1] * i) % 998244353;
  map<long long, map<long long, long long> > make_pair;
  for (long long i = 0; i < n; i++) {
    make_pair[second[i].first][second[i].second]++;
  }
  long long tot = 1;
  long long mn = 1;
  long long prev = -1000000000000000LL;
  for (map<long long, map<long long, long long> >::iterator it =
           make_pair.begin();
       it != make_pair.end(); ++it) {
    long long temp = 1;
    long long t1 = 0;
    for (map<long long, long long>::iterator it1 = (it->second).begin();
         it1 != (it->second).end(); it1++) {
      temp *= first[it1->second];
      t1 += it1->second;
      t1 %= 998244353;
      temp %= 998244353;
    }
    t1 = first[t1];
    map<long long, long long>::iterator it2 = (it->second).begin();
    if (it2->first >= prev)
      mn *= temp;
    else
      mn = 0;
    it2 = std::prev((it->second).end());
    prev = it2->first;
    mn %= 998244353;
    tot *= t1;
    tot %= 998244353;
  }
  make_pair.clear();
  long long tot1 = 1;
  for (long long i = 0; i < n; i++) {
    make_pair[second[i].second][second[i].first]++;
  }
  for (map<long long, map<long long, long long> >::iterator it =
           make_pair.begin();
       it != make_pair.end(); ++it) {
    long long temp = 1;
    long long t1 = 0;
    for (map<long long, long long>::iterator it1 = (it->second).begin();
         it1 != (it->second).end(); it1++) {
      temp *= first[it1->second];
      t1 += it1->second;
      t1 %= 998244353;
      temp %= 998244353;
    }
    t1 = first[t1];
    tot1 *= t1;
    tot1 %= 998244353;
  }
  long long ans = first[n];
  ans -= tot + tot1 - mn;
  ans %= 998244353;
  ans += 998244353;
  ans %= 998244353;
  cout << ans << '\n';
}
