#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int MOD = 998244353;
  long long int n;
  cin >> n;
  pair<long long int, long long int> arr[n];
  for (long long int i = 0; i < n; ++i) cin >> arr[i].first >> arr[i].second;
  long long int fact[n + 1];
  fact[0] = 1;
  for (long long int i = 1; i <= n; ++i) {
    fact[i] = (fact[i - 1] * i + MOD) % MOD;
  }
  map<long long int, long long int> count1;
  map<long long int, long long int> count2;
  map<pair<long long int, long long int>, long long int> count3;
  for (long long int i = 0; i < n; ++i) {
    count1[arr[i].first]++;
    count2[arr[i].second]++;
    count3[make_pair(arr[i].first, arr[i].second)]++;
  }
  long long int n1 = 1, n2 = 1, n3 = 1;
  map<long long int, long long int>::iterator it;
  for (it = count1.begin(); it != count1.end(); ++it) {
    n1 = (n1 * (fact[it->second]) + MOD) % MOD;
  }
  for (it = count2.begin(); it != count2.end(); ++it) {
    n2 = (n2 * (fact[it->second]) + MOD) % MOD;
  }
  map<pair<long long int, long long int>, long long int>::iterator itd;
  long long int last2 = -1;
  for (itd = count3.begin(); itd != count3.end(); ++itd) {
    n3 = (n3 * (fact[itd->second]) + MOD) % MOD;
    if (itd->first.second < last2) {
      n3 = 0;
      break;
    }
    last2 = itd->first.second;
  }
  long long int ans = (fact[n] - n1 - n2 + n3 + 2 * MOD) % MOD;
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int MOD = 998244353;
  long long int n;
  cin >> n;
  pair<long long int, long long int> arr[n];
  for (long long int i = 0; i < n; ++i) cin >> arr[i].first >> arr[i].second;
  long long int fact[n + 1];
  fact[0] = 1;
  for (long long int i = 1; i <= n; ++i) {
    fact[i] = (fact[i - 1] * i + MOD) % MOD;
  }
  map<long long int, long long int> count1;
  map<long long int, long long int> count2;
  map<pair<long long int, long long int>, long long int> count3;
  for (long long int i = 0; i < n; ++i) {
    count1[arr[i].first]++;
    count2[arr[i].second]++;
    count3[make_pair(arr[i].first, arr[i].second)]++;
  }
  long long int n1 = 1, n2 = 1, n3 = 1;
  map<long long int, long long int>::iterator it;
  for (it = count1.begin(); it != count1.end(); ++it) {
    n1 = (n1 * (fact[it->second]) + MOD) % MOD;
  }
  for (it = count2.begin(); it != count2.end(); ++it) {
    n2 = (n2 * (fact[it->second]) + MOD) % MOD;
  }
  map<pair<long long int, long long int>, long long int>::iterator itd;
  long long int last2 = -1;
  for (itd = count3.begin(); itd != count3.end(); ++itd) {
    n3 = (n3 * (fact[itd->second]) + MOD) % MOD;
    if (itd->first.second < last2) {
      n3 = 0;
      break;
    }
    last2 = itd->first.second;
  }
  long long int ans = (fact[n] - n1 - n2 + n3 + 2 * MOD) % MOD;
  cout << ans;
  return 0;
}

```