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;
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  out << "{";
  for (int i = (0); i < (v.size()); i++) {
    if (i > 0) out << ", ";
    out << v[i];
  }
  return out << "}";
}
template <typename A, typename B>
ostream &operator<<(ostream &out, pair<A, B> &p) {
  return out << "(" << p.first << ", " << p.second << ")";
}
const int N = 3e5 + 5;
const int inf = 1e9 + 5;
const long long mod = 998244353;
pair<int, int> a[N];
long long f[N];
int n;
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) { return ((x - y) % mod + mod) % mod; }
long long mul(long long x, long long y) { return (x * y) % mod; }
long long one() {
  sort(a, a + n);
  long long tt = 1, cnt = 0, bad = 1;
  for (int i = (0); i < (n); i++) {
    cnt++;
    if (a[i].first != a[i + 1].first) {
      tt = mul(tt, f[cnt]), cnt = 0;
    }
  }
  map<int, int> m;
  for (int i = (0); i < (n); i++) {
    m[a[i].second]++;
    if (a[i].first != a[i + 1].first) {
      for (auto it = m.begin(); it != m.end(); it++) {
        bad = mul(bad, f[it->second]);
      }
      m.clear();
    }
  }
  for (int i = (0); i < (n - 1); i++)
    if (a[i].second > a[i + 1].second) bad = 0;
  return sub(tt, bad);
}
long long both() {
  sort(a, a + n);
  long long res = 1;
  map<int, int> m;
  for (int i = (0); i < (n); i++) {
    m[a[i].second]++;
    if (a[i].first != a[i + 1].first) {
      for (auto it = m.begin(); it != m.end(); it++) {
        res = mul(res, f[it->second]);
      }
      m.clear();
    }
  }
  for (int i = (0); i < (n - 1); i++)
    if (a[i].second > a[i + 1].second) return 0;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  f[0] = 1;
  for (int i = (1); i < (3e5 + 1); i++) {
    f[i] = mul(i, f[i - 1]);
  }
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> a[i].first >> a[i].second;
  a[n].first = a[n].second = inf;
  long long tt = f[n], o1 = one();
  for (int i = (0); i < (n); i++) swap(a[i].first, a[i].second);
  long long o2 = one();
  long long b = both();
  tt = sub(tt, o1);
  tt = sub(tt, o2);
  tt = sub(tt, b);
  cout << tt << "\n";
  return 0;
}
