#include <bits/stdc++.h>
using namespace std;
long long int arr1[300005] = {0}, arr2[300005] = {0}, factorial[300005];
void solve() {
  long long int i;
  factorial[0] = 1;
  for (i = 1; i < 300005; i++)
    factorial[i] = (factorial[i - 1] * i) % 998244353;
}
int main() {
  long long int n, i, a, b;
  cin >> n;
  if (n == 1)
    cout << 0;
  else {
    solve();
    vector<pair<long long int, long long int> > v;
    for (i = 0; i < n; i++) {
      cin >> a >> b;
      arr1[a]++;
      arr2[b]++;
      v.push_back(make_pair(a, b));
    }
    long long int ans1 = 1, ans2 = 1, count = 1, flag = 0;
    for (i = 0; i < 300005; i++) ans1 = (ans1 * factorial[arr1[i]]) % 998244353;
    for (i = 0; i < 300005; i++) ans2 = (ans2 * factorial[arr2[i]]) % 998244353;
    sort(v.begin(), v.end());
    ans1 = (ans1 + ans2) % 998244353;
    ans2 = 1;
    for (i = 1; i < n; i++) {
      if (v[i].second < v[i - 1].second) {
        flag = 1;
        break;
      }
      if (v[i].first == v[i - 1].first && v[i].second == v[i - 1].second)
        count++;
      else {
        ans2 = (ans2 * (factorial[count])) % 998244353;
        count = 1;
      }
    }
    ans2 = (ans2 * (factorial[count])) % 998244353;
    if (flag == 0) ans1 = (ans1 - ans2 + 998244353) % 998244353;
    ans1 = (factorial[n] - ans1 + 998244353) % 998244353;
    cout << ans1;
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int arr1[300005] = {0}, arr2[300005] = {0}, factorial[300005];
void solve() {
  long long int i;
  factorial[0] = 1;
  for (i = 1; i < 300005; i++)
    factorial[i] = (factorial[i - 1] * i) % 998244353;
}
int main() {
  long long int n, i, a, b;
  cin >> n;
  if (n == 1)
    cout << 0;
  else {
    solve();
    vector<pair<long long int, long long int> > v;
    for (i = 0; i < n; i++) {
      cin >> a >> b;
      arr1[a]++;
      arr2[b]++;
      v.push_back(make_pair(a, b));
    }
    long long int ans1 = 1, ans2 = 1, count = 1, flag = 0;
    for (i = 0; i < 300005; i++) ans1 = (ans1 * factorial[arr1[i]]) % 998244353;
    for (i = 0; i < 300005; i++) ans2 = (ans2 * factorial[arr2[i]]) % 998244353;
    sort(v.begin(), v.end());
    ans1 = (ans1 + ans2) % 998244353;
    ans2 = 1;
    for (i = 1; i < n; i++) {
      if (v[i].second < v[i - 1].second) {
        flag = 1;
        break;
      }
      if (v[i].first == v[i - 1].first && v[i].second == v[i - 1].second)
        count++;
      else {
        ans2 = (ans2 * (factorial[count])) % 998244353;
        count = 1;
      }
    }
    ans2 = (ans2 * (factorial[count])) % 998244353;
    if (flag == 0) ans1 = (ans1 - ans2 + 998244353) % 998244353;
    ans1 = (factorial[n] - ans1 + 998244353) % 998244353;
    cout << ans1;
  }
}

```