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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n], b[n];
  pair<int, int> ab[n];
  long long facts[n + 1], mod = 998244353;
  facts[0] = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    ab[i] = pair<int, int>(a[i], b[i]);
    facts[i + 1] = (facts[i] * (i + 1)) % mod;
  }
  sort(a, a + n);
  long long aord = 1;
  for (int i = 0; i < n; i++) {
    int rep = 1;
    while (++i < n && a[i] == a[i - 1]) rep++;
    i--;
    aord = (aord * facts[rep]) % mod;
  }
  sort(b, b + n);
  long long bord = 1;
  for (int i = 0; i < n; i++) {
    int rep = 1;
    while (++i < n && b[i] == b[i - 1]) rep++;
    i--;
    bord = (bord * facts[rep]) % mod;
  }
  sort(ab, ab + n);
  bool cord = true;
  for (int i = 1; i < n && cord; i++)
    cord =
        (ab[i - 1].first <= ab[i].first) && (ab[i - 1].second <= ab[i].second);
  if (cord) {
    long long abord = 1;
    for (int i = 0; i < n; i++) {
      int rep = 1;
      while (++i < n && ab[i] == ab[i - 1]) rep++;
      i--;
      abord = (abord * facts[rep]) % mod;
    }
    cout << (mod + (facts[n] - aord - bord + abord) % mod) % mod;
  } else
    cout << (mod + (facts[n] - aord - bord) % mod) % mod << '\n';
  return 0;
}
