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 MOD = 998244353;
int MX = 1e9;
long double pi = 3.1415926535;
long long int n;
pair<long long int, long long int> a[300005];
pair<long long int, long long int> b[300005];
long long int fact[300005];
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  fact[0] = fact[1] = 1;
  for (long long int i = 2; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
    b[i].first = a[i].second;
    b[i].second = a[i].first;
  }
  sort(a + 1, a + n + 1);
  sort(b + 1, b + n + 1);
  long long int res = 0;
  long long int ans = 1;
  for (int i = 1; i <= n; i++) {
    int st = a[i].first;
    long long int temp = 0;
    while (a[i].first == st) {
      i++;
      temp++;
    }
    i--;
    ans = (ans * fact[temp]) % MOD;
  }
  res += ans;
  ans = 1;
  for (int i = 1; i <= n; i++) {
    int st = b[i].first;
    long long int temp = 0;
    while (b[i].first == st) {
      i++;
      temp++;
    }
    i--;
    ans = (ans * fact[temp]) % MOD;
  }
  res += ans;
  res %= MOD;
  ans = 1;
  for (int i = 1; i <= n; i++) {
    int st1 = a[i].first;
    int st2 = a[i].second;
    long long int temp = 0;
    while (a[i].second == st2 && a[i].first == st1) {
      i++;
      temp++;
    }
    i--;
    ans = (ans * fact[temp]) % MOD;
  }
  bool ff = 0;
  for (int i = 2; i <= n; i++) {
    if (a[i].second < a[i - 1].second) ff = 1;
  }
  if (ff == 0) res = (res - ans + MOD) % MOD;
  res = (MOD + fact[n] - res) % MOD;
  cout << res;
  return 0;
}
