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 inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
double eps = 1e-6;
const long long int mod = 998244353;
int a, n;
string str, ptr;
int arr[2001];
long long dp[2001][2001];
bool compare1(pair<int, int> p1, pair<int, int> p2) {
  return p1.first < p2.first;
}
bool compare2(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
bool compare3(pair<int, int> p1, pair<int, int> p2) {
  return p1.first <= p2.first && p1.second <= p2.second;
}
int main() {
  long long int permu[400000];
  permu[1] = 1;
  for (int i = 2; i < 400000; i++) {
    permu[i] = (i * permu[i - 1]) % mod;
  }
  long long int a, b, c, d;
  int n;
  cin >> n;
  vector<pair<int, int> > values1(n, {0, 0});
  for (int i = 0; i < n; ++i) {
    cin >> values1[i].first >> values1[i].second;
  }
  sort(values1.begin(), values1.end(), compare1);
  long long int totalsum1 = 1LL, totalsum2 = 1LL, totalsum = 1LL;
  int counter = 1;
  for (int i = 1; i < n; i++) {
    if (values1[i].first == values1[i - 1].first) {
      counter++;
    } else {
      totalsum1 = (totalsum1 * permu[counter]) % mod;
      counter = 1;
    }
  }
  totalsum1 = (totalsum1 * permu[counter]) % mod;
  sort(values1.begin(), values1.end(), compare2);
  counter = 1;
  for (int i = 1; i < n; i++) {
    if (values1[i].second == values1[i - 1].second) {
      counter++;
    } else {
      totalsum2 = (totalsum2 * permu[counter]) % mod;
      counter = 1;
    }
  }
  totalsum2 = (totalsum2 * permu[counter]) % mod;
  totalsum = 0LL;
  counter = 1;
  totalsum += (totalsum1 + totalsum2) % mod;
  sort(values1.begin(), values1.end());
  long long int totalsum3 = 1LL;
  for (int i = 1; i < n; i++) {
    if (values1[i].first >= values1[i - 1].first &&
        values1[i].second >= values1[i - 1].second) {
    } else {
      totalsum3 = 0ll;
    }
    if (values1[i].first == values1[i - 1].first &&
        values1[i].second == values1[i - 1].second) {
      counter++;
    } else {
      totalsum3 = (totalsum3 * permu[counter]) % mod;
      counter = 1;
    }
  }
  totalsum3 = (totalsum3 * permu[counter]) % mod;
  totalsum -= totalsum3;
  totalsum += mod;
  totalsum %= mod;
  totalsum = permu[n] - totalsum + mod;
  totalsum %= mod;
  printf("%lld\n", totalsum);
}
