#include <bits/stdc++.h>
using namespace std;
constexpr int mod = 998244353, Size = 300000;
pair<int, int> s[Size];
bool cmpf(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
bool cmps(pair<int, int> a, pair<int, int> b) {
  if (a.second == b.second) {
    return a.first < b.first;
  } else {
    return a.second < b.second;
  }
}
int main() {
  int n;
  long long int ans = 1, comf = 1, coms = 1, coma = 1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i].first >> s[i].second;
  }
  for (int m = 1; m <= n; m++) {
    ans *= m;
    ans %= mod;
  }
  sort(s, s + n, cmpf);
  for (int i = 0; i < n;) {
    int tmp = s[i].first, c = 0;
    while (s[i].first == tmp) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      comf *= j;
      comf %= mod;
    }
  }
  sort(s, s + n, cmps);
  for (int i = 0; i < n;) {
    int tmp = s[i].second, c = 0;
    while (s[i].second == tmp) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      coms *= j;
      coms %= mod;
    }
  }
  for (int i = 0; i < n;) {
    int tmp1 = s[i].first, tmp2 = s[i].second, c = 0;
    while (s[i].first == tmp1 && s[i].second == tmp2) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      coma *= j;
      coma %= mod;
    }
    if (s[i].first < s[i - 1].first && i < n) {
      coma = 0;
    }
  }
  ans -= (comf + coms - coma);
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to 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;
constexpr int mod = 998244353, Size = 300000;
pair<int, int> s[Size];
bool cmpf(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
bool cmps(pair<int, int> a, pair<int, int> b) {
  if (a.second == b.second) {
    return a.first < b.first;
  } else {
    return a.second < b.second;
  }
}
int main() {
  int n;
  long long int ans = 1, comf = 1, coms = 1, coma = 1;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i].first >> s[i].second;
  }
  for (int m = 1; m <= n; m++) {
    ans *= m;
    ans %= mod;
  }
  sort(s, s + n, cmpf);
  for (int i = 0; i < n;) {
    int tmp = s[i].first, c = 0;
    while (s[i].first == tmp) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      comf *= j;
      comf %= mod;
    }
  }
  sort(s, s + n, cmps);
  for (int i = 0; i < n;) {
    int tmp = s[i].second, c = 0;
    while (s[i].second == tmp) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      coms *= j;
      coms %= mod;
    }
  }
  for (int i = 0; i < n;) {
    int tmp1 = s[i].first, tmp2 = s[i].second, c = 0;
    while (s[i].first == tmp1 && s[i].second == tmp2) {
      c++;
      i++;
      if (i >= n) break;
    }
    for (int j = 1; j <= c; j++) {
      coma *= j;
      coma %= mod;
    }
    if (s[i].first < s[i - 1].first && i < n) {
      coma = 0;
    }
  }
  ans -= (comf + coms - coma);
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}

```