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 n, ans, k1, k2, k3, i, j;
map<long long, int> m1, m;
vector<int> v, v1;
pair<long long, long long> a[3 * 100500];
long long fact(long long n) {
  long long i, k = 1;
  for (i = 2; i <= n; i++) k *= i, k %= 998244353;
  return k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  ans = fact(n);
  k1 = k2 = 1;
  for (i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
    m[a[i].first]++;
    m1[a[i].second]++;
    if (m[a[i].first] == 1) v.push_back(a[i].first);
    if (m1[a[i].second] == 1) v1.push_back(a[i].second);
  }
  for (i = 0; i < v.size(); i++) k1 *= fact(m[v[i]]), k1 %= 998244353;
  for (i = 0; i < v1.size(); i++) k2 *= fact(m1[v1[i]]), k2 %= 998244353;
  k3 = 1;
  sort(a + 1, a + 1 + n);
  bool f = 0;
  for (i = 2; i <= n; i++)
    if (a[i].second < a[i - 1].second) f = 1;
  if (f)
    cout << (ans - k1 - k2 + 2 * 998244353) % 998244353;
  else {
    i = 1;
    while (i <= n) {
      j = i + 1;
      while (j <= n && a[j].first == a[j - 1].first &&
             a[j].second == a[j - 1].second)
        j++;
      k3 *= fact(j - i);
      k3 %= 998244353;
      i = j;
    }
    cout << ((ans - k1 - k2 + 2 * 998244353) % 998244353 + k3) % 998244353;
  }
}
