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;
const unsigned long long p = 998244353;
int main() {
  unsigned long long n, i, x, y, a = 1, b = 1, c = 1, cnt = 1, pre;
  cin >> n;
  unsigned long long fact[n + 1];
  fact[0] = 1;
  for (i = 1; i <= n; i++) fact[i] = (fact[i - 1] % p * i % p) % p;
  vector<pair<unsigned long long, unsigned long long> > v1, v2;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    v1.push_back({x, y});
    v2.push_back({y, x});
  }
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  pre = v1[0].first;
  for (i = 1; i < n; i++) {
    if (v1[i].first == pre)
      cnt++;
    else {
      a = (a % p * fact[cnt] % p) % p;
      pre = v1[i].first;
      cnt = 1;
    }
  }
  a = (a % p * fact[cnt] % p) % p;
  pre = v2[0].first;
  cnt = 1;
  for (i = 1; i < n; i++) {
    if (v2[i].first == pre)
      cnt++;
    else {
      b = (b % p * fact[cnt] % p) % p;
      pre = v2[i].first;
      cnt = 1;
    }
  }
  b = (b % p * fact[cnt] % p) % p;
  pair<unsigned long long, unsigned long long> pr = v1[0];
  cnt = 1;
  for (i = 1; i < n; i++) {
    if (v1[i].second < pr.second) {
      c = 0;
      break;
    }
    if (v1[i] == pr)
      cnt++;
    else {
      c = (c % p * fact[cnt] % p) % p;
      pr = v1[i];
      cnt = 1;
    }
  }
  c = (c % p * fact[cnt] % p) % p;
  unsigned long long ans = (fact[n] - a - b + c + p + p) % p;
  cout << ans;
  return 0;
}
