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;
pair<long long, long long> a[300000];
long long f[300001];
long long mod = 998244353;
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  f[0] = 1;
  for (long long i = 1; i <= n; ++i) {
    f[i] = (f[i - 1] * i) % mod;
    cin >> a[i - 1].first >> a[i - 1].second;
  }
  sort(a, a + n);
  pair<long long, long long> p = a[0];
  long long cp = 0, ca = 0, rp = 1, ra = 1;
  for (long long i = 0; i <= n; ++i) {
    if (i < n && a[i].second < p.second) rp = 0;
    if (i == n || p != a[i]) {
      if (i == n || p.first != a[i].first) {
        ra = (ra * f[ca]) % mod;
        ca = 1;
      } else
        ++ca;
      rp = (rp * f[cp]) % mod;
      cp = 1;
      if (i < n) p = a[i];
    } else {
      ++cp;
      ++ca;
    }
    swap(a[i].first, a[i].second);
  }
  sort(a, a + n);
  long long cb = 0, rb = 1;
  p = a[0];
  for (long long i = 0; i <= n; ++i) {
    if (i == n || p.first != a[i].first) {
      rb = (rb * f[cb]) % mod;
      cb = 1;
      if (i < n) p = a[i];
    } else
      ++cb;
  }
  long long r = (f[n] + rp) % mod;
  r = (r + mod - ra) % mod;
  r = (r + mod - rb) % mod;
  cout << r << endl;
  return 0;
}
