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;
bool cmp(pair<long long int, long long int> p1,
         pair<long long int, long long int> p2) {
  if (p1.first != p2.first)
    return (p1.first < p2.first);
  else
    return (p1.second < p2.second);
}
bool cmp2(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  return (p1.second < p2.second);
}
int main() {
  long long int n, m = 998244353;
  cin >> n;
  pair<long long int, long long int> arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i].first >> arr[i].second;
  long long int factorial[n + 1];
  factorial[0] = 1;
  for (int i = 1; i <= n; i++) factorial[i] = (factorial[i - 1] * i) % m;
  sort(arr, arr + n, cmp);
  long long int temp = 1, temp2 = 1, temp3 = 1;
  long long int i = 0;
  while (i < n) {
    long long int el = arr[i].first;
    long long int cnt = 0;
    while (i < n && arr[i].first == el) {
      i++;
      cnt++;
    }
    temp = (temp * factorial[cnt]) % m;
  }
  i = 0;
  sort(arr, arr + n, cmp2);
  while (i < n) {
    long long int el = arr[i].second;
    long long int cnt = 0;
    while (i < n && arr[i].second == el) {
      i++;
      cnt++;
    }
    temp2 = (temp2 * factorial[cnt]) % m;
  }
  int flag = 0;
  sort(arr, arr + n, cmp);
  for (int i = 1; i < n; i++) {
    if (arr[i].first < arr[i - 1].first || arr[i].second < arr[i - 1].second) {
      flag = 1;
      break;
    }
  }
  if (flag == 0) {
    i = 0;
    while (i < n) {
      long long int el = arr[i].first;
      long long int el2 = arr[i].second;
      long long int cnt = 0;
      while (i < n && arr[i].first == el && arr[i].second == el2) {
        i++;
        cnt++;
      }
      temp3 = (temp3 * factorial[cnt]) % m;
    }
    cout << ((factorial[n] - ((temp2 + temp - temp3 + m) % m) + m) % m);
  } else {
    cout << ((factorial[n] - ((temp2 + temp) % m) + m) % m);
  }
  return 0;
}
