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. 
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;

public class Main 
{
	public static long Mod = 998244353;
	public static class Pair 
	{
		public long first;
		public long second;
		Pair(){first = 0; second = 0;}
		Pair(int first, int second){this.first = first; this.second = second;}
	}
	public static long fact(long x)
	{
		long res = 1;
		for(long val = 1 ; val <= x ; val++)
			res = (res * val) % Mod;
		return res;
	}
	public static boolean isSorted(ArrayList<Pair> list)
	{
		for(int i = 1 ; i < list.size(); i++)
			if(list.get(i).first < list.get(i - 1).first || list.get(i).second < list.get(i - 1).second)
				return false;
		return true;
	}
	public static void main(String[] args) 
	{
		Scanner scanner = new Scanner(System.in);
		long n = scanner.nextInt();
		long[] freq1 = new long[(int)n + 1];
		long[] freq2 = new long[(int)n + 1];
		ArrayList<Pair> list = new ArrayList<>();
		for(int i = 0, a, b ; i < n ; i++)
		{
			a = scanner.nextInt();
			b = scanner.nextInt();
			list.add(new Pair(a, b));
			freq1[a]++;
			freq2[b]++;
		}
		long sub1 = 1, sub2 = 1, total = fact(n);
		for(int i = 1 ; i <= n ; i++)
			sub1 = (sub1 * fact(freq1[i])) % Mod;
		for(int i = 1 ; i <= n ; i++)
			sub2 = (sub2 * fact(freq2[i])) % Mod;
		
		Collections.sort(list, new Comparator<Pair>() {
			public int compare(Pair p1, Pair p2) {
				if(p1.first < p2.first) return -1;
				else if(p1.first == p2.first)
				{
					if(p1.second < p2.second) return -1;
					else if(p1.second == p2.second) return 0;
					else return 1;
				}
				else return 1;
			}
		});
		if(isSorted(list))
		{
			long add = 1, cnt = 1;
			for(int i = 1 ; i < list.size(); i++)
			{
				if(list.get(i).first == list.get(i - 1).first && list.get(i).second == list.get(i - 1).second)
					++cnt;
				else
				{
					add = (add * fact(cnt)) % Mod;
					cnt = 1;
				}
			}
			add = (add * fact(cnt)) % Mod;
			total = total + add;
		}
		System.out.println((total - sub1 - sub2 + 2 * Mod) % Mod);
		scanner.close();
	}
}
