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;
long long int fact(long long int n) {
  long long int ans = 1;
  for (long long int j = 1; j <= n; j++) {
    ans *= j;
    ans = ans % 998244353;
  }
  return ans;
}
int32_t main() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> ele;
  map<long long int, long long int> sorted_x;
  map<long long int, long long int> sorted_y;
  map<pair<long long int, long long int>, long long int> pcount;
  for (long long int j = 0; j < n; j++) {
    long long int a, b;
    cin >> a >> b;
    ele.push_back({a, b});
    pcount[{a, b}] += 1;
    sorted_x[a] += 1;
    sorted_y[b] += 1;
  }
  sort(ele.begin(), ele.end());
  vector<long long int> x(n);
  vector<long long int> y(n);
  for (long long int j = 0; j < n; j++) {
    x[j] = ele[j].first;
    y[j] = ele[j].second;
  }
  if (is_sorted(y.begin(), y.end())) {
    long long int ans = fact(n);
    long long int to_sub = 1;
    for (auto it = sorted_x.begin(); it != sorted_x.end(); it++) {
      to_sub *= fact(it->second);
      to_sub = to_sub % 998244353;
    }
    ans = ans - to_sub;
    if (ans < 0) {
      ans += 998244353;
    }
    long long int to_sub_2 = 1;
    for (auto it = sorted_y.begin(); it != sorted_y.end(); it++) {
      to_sub_2 *= fact(it->second);
      to_sub_2 = to_sub_2 % 998244353;
    }
    long long int to_add = 1;
    for (auto it = pcount.begin(); it != pcount.end(); it++) {
      to_add *= fact(it->second);
      to_add = to_add % 998244353;
    }
    ans -= to_sub_2;
    if (ans < 0) {
      ans += 998244353;
    }
    ans += to_add;
    ans = ans % 998244353;
    cout << ans << "\n";
  } else {
    long long int to_sub_1 = 1;
    for (auto it = sorted_x.begin(); it != sorted_x.end(); it++) {
      to_sub_1 *= fact(it->second);
      to_sub_1 = to_sub_1 % 998244353;
    }
    long long int to_sub_2 = 1;
    for (auto it = sorted_y.begin(); it != sorted_y.end(); it++) {
      to_sub_2 *= fact(it->second);
      to_sub_2 = to_sub_2 % 998244353;
    }
    long long int ans = fact(n);
    ans -= to_sub_1;
    if (ans < 0) {
      ans += 998244353;
    } else {
      ans = ans % 998244353;
    }
    ans -= to_sub_2;
    if (ans < 0) {
      ans += 998244353;
    } else {
      ans = ans % 998244353;
    }
    cout << ans << "\n";
  }
  return 0;
}
