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 <class T>
void print(T &x) {
  cout << x;
}
template <class T>
void print(vector<T> &v) {
  cout << "{";
  if ((v).size()) print(v[0]);
  for (int i = 1; i < (v).size(); i++) cout << ", ", print(v[i]);
  cout << "}\n";
}
const int N = 105;
int q, k, t, m, n, cs = 1;
const long long mod = 998244353;
long long fact(long long num) {
  return num == 0 ? 1 : num * fact(num - 1) % mod;
}
inline void relax(long long &res) {
  res %= mod;
  res += mod;
  res %= mod;
}
long long ckBoth(vector<pair<int, int> > &pr) {
  sort((pr).begin(), (pr).end());
  for (int i = 1; i < (pr).size(); ++i) {
    if (pr[i].second < pr[i - 1].second) return 0;
  }
  long long res = 1;
  for (int i = 0; i < (pr).size();) {
    int j = i + 1;
    while (pr[i] == pr[j] and j < (pr).size()) j++;
    res = res * fact(j - i);
    relax(res);
    i = j;
  }
  return res;
}
void HalfDead() {
  vector<int> a, b;
  vector<pair<int, int> > pr;
  int x, y;
  cin >> n;
  for (int i = 0; i < (int)n; ++i) {
    cin >> x >> y;
    a.push_back(x);
    b.push_back(y);
    pr.push_back({x, y});
  }
  long long res = fact(n);
  relax(res);
  sort((a).begin(), (a).end());
  sort((b).begin(), (b).end());
  long long r = 1;
  for (int i = 0; i < (a).size();) {
    int j = i + 1;
    while (a[i] == a[j] and j < (a).size()) j++;
    r *= fact(j - i);
    relax(r);
    i = j;
  }
  res -= r;
  relax(res);
  r = 1;
  for (int i = 0; i < (b).size();) {
    int j = i + 1;
    while (b[i] == b[j] and j < (b).size()) j++;
    r *= fact(j - i);
    relax(r);
    i = j;
  }
  res -= r;
  relax(res);
  res += ckBoth(pr);
  relax(res);
  cout << res << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(0);
  int tc = 1;
  while (tc--) {
    HalfDead();
  }
  return 0;
}
