#include <bits/stdc++.h>
using namespace std;
long long p = 998244353;
int a[100050 * 3] = {0}, b[100050 * 3] = {0}, c[100050 * 3] = {0};
long long ft[100050 * 3] = {1, 1};
pair<int, int> ip[100050 * 3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    ip[i] = {x, y};
    a[x]++, b[y]++;
    if (x == y) c[x]++;
  }
  for (long long i = 2; i <= n; i++) {
    ft[i] = ft[i - 1] * i % p;
  }
  long long as = 1, bs = 1;
  for (int i = 1; i <= n; i++) {
    as = as * ft[a[i]] % p;
    bs = bs * ft[b[i]] % p;
  }
  sort(ip + 1, ip + n + 1);
  int flag = 1;
  for (int i = 2; i <= n; i++) {
    if (ip[i].second < ip[i - 1].second) flag = 0;
  }
  if (flag == 0) {
    cout << (ft[n] - as - bs + p + p) % p << '\n';
  } else {
    long long cs = 1, now = 1;
    for (int i = 2; i <= n; i++) {
      if (ip[i].first != ip[i - 1].first || ip[i].second != ip[i - 1].second) {
        cs = cs * ft[now] % p;
        now = 1;
      } else {
        now++;
      }
    }
    cs = cs * ft[now] % p;
    cout << (ft[n] - as - bs + p + p + cs) % p << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long p = 998244353;
int a[100050 * 3] = {0}, b[100050 * 3] = {0}, c[100050 * 3] = {0};
long long ft[100050 * 3] = {1, 1};
pair<int, int> ip[100050 * 3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    ip[i] = {x, y};
    a[x]++, b[y]++;
    if (x == y) c[x]++;
  }
  for (long long i = 2; i <= n; i++) {
    ft[i] = ft[i - 1] * i % p;
  }
  long long as = 1, bs = 1;
  for (int i = 1; i <= n; i++) {
    as = as * ft[a[i]] % p;
    bs = bs * ft[b[i]] % p;
  }
  sort(ip + 1, ip + n + 1);
  int flag = 1;
  for (int i = 2; i <= n; i++) {
    if (ip[i].second < ip[i - 1].second) flag = 0;
  }
  if (flag == 0) {
    cout << (ft[n] - as - bs + p + p) % p << '\n';
  } else {
    long long cs = 1, now = 1;
    for (int i = 2; i <= n; i++) {
      if (ip[i].first != ip[i - 1].first || ip[i].second != ip[i - 1].second) {
        cs = cs * ft[now] % p;
        now = 1;
      } else {
        now++;
      }
    }
    cs = cs * ft[now] % p;
    cout << (ft[n] - as - bs + p + p + cs) % p << '\n';
  }
  return 0;
}

```