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.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
	static final long mod=998244353;
	public static void main(String[] args) throws Exception
	{
		FastReader in=new FastReader();
		PrintWriter pw=new PrintWriter(System.out);
		int n=in.nextInt();
		Integer[][] arr=new Integer[n][2];
		long[] fact=new long[n+1];
		fact[0]=1;
		for(int i=0;i<n;i++)
		{
			fact[i+1]=(fact[i]*(i+1))%mod;
			arr[i][0]=in.nextInt();
			arr[i][1]=in.nextInt();
		}
		Arrays.sort(arr,(Integer[] ar,Integer[] br)-> Integer.compare(ar[0],br[0]));
		long a=1,b=1;
		int c=0,prev=-1;
		for(int i=0;i<n;i++)
		{
			if(prev!=arr[i][0])
			{
				a=(a*fact[c])%mod;
				prev=arr[i][0];
				c=1;
			}
			else
				c++;
		}
		a=(a*fact[c])%mod;
		Arrays.sort(arr,(Integer[] ar,Integer[] br)-> Integer.compare(ar[1],br[1]));
		c=0;
		prev=-1;
		for(int i=0;i<n;i++)
		{
			if(prev!=arr[i][1])
			{
				b=(b*fact[c])%mod;
				prev=arr[i][1];
				c=1;
			}
			else
				c++;
		}
		b=(b*fact[c])%mod;
		Arrays.sort(arr,(Integer[] ar,Integer[] br)->
		{
			if(Integer.compare(ar[0],br[0])==0)
				return Integer.compare(ar[1],br[1]);
			return Integer.compare(ar[0],br[0]);
		});
		long com=1;
		c=0;
		int prev1=-1,prev2=-1;
		for(int i=0;i<n;i++)
		{
			if(prev1!=arr[i][0] && prev2>arr[i][1])
			{
				com=0;
				break;
			}
			if(prev1!=arr[i][0] || prev2!=arr[i][1])
			{
				com=(com*fact[c])%mod;
				prev1=arr[i][0];
				prev2=arr[i][1];
				c=1;
			}
			else
				c++;
		}
		com=(com*fact[c])%mod;
		long ans=fact[n]-(a+b-com)%mod;
		while(ans<0)
			ans+=mod;
		pw.print(ans);
		pw.flush();
	}
}

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
	    br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() throws IOException
	{
	    if(st==null || !st.hasMoreElements())
	    {
	        st=new StringTokenizer(br.readLine());
	    }
	    return st.nextToken();
	}

	public int nextInt() throws IOException
	{
	    return Integer.parseInt(next());
	}

	public long nextLong() throws IOException
	{
	    return Long.parseLong(next());
	}
}