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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.stream.Collectors;


public class D{
	
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
	static int l,r;
	static long sum = 0L;
	static void maxSum(long a[])
	{
		long max_so_far = Long.MIN_VALUE, 
		        max_ending_here = 0;
		int start = 0, end = 0, s = 0; 
		int size = a.length; 
		        for (int i = 0; i < size; i++)  
		        { 
		            max_ending_here += a[i]; 
		  
		            if (max_so_far < max_ending_here)  
		            { 
		                max_so_far = max_ending_here; 
		                start = s; 
		                end = i; 
		            } 
		  
		            if (max_ending_here < 0)  
		            { 
		                max_ending_here = 0; 
		                s = i + 1; 
		            } 
		        }
		        if(max_so_far > 0)
		        {
		        	l = start;
		        	r = end;
		        }
		        else
		        {
		        	l = -1;
		        	r = -1;
		        }
		        sum = max_so_far;
	}
	
	static class Pair
	{
		int a;
		int b;
		Pair(int a,int b)
		{
			this.a = a;
			this.b = b;
		}
	}
	
	static long mod;
	static long fact(long n)
	{	
		long f = 1L;
		
		for(long i = 2; i <= n; i++)
		{
			f = (f*i)%mod;
		}
		return f;
	}
	public static void main(String[] args) 
	{
		OutputStream outputStream = System.out;
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(outputStream);
        
        int n = sc.nextInt();
        Pair[] pr = new Pair[n];
        mod = 998244353;
        for(int i = 0; i < n; i++)
        {
        	pr[i] = new Pair(sc.nextInt(),sc.nextInt());
        }
        
        
        Arrays.sort(pr,new Comparator<Pair>() { 
            @Override public int compare(Pair p1, Pair p2) 
            { 	
            	if(p1.a != p2.a)
                return p1.a - p2.a;  //ascending , for descending -> p2.a - p1.a
            	
            	else
            	return p1.b - p2.b;
            } 
        });
        
        long count = 1;
        long ans = 1L;
        for(int i = 1; i < n; i++)
        {
        	if(pr[i-1].a == pr[i].a)
        	{
        		count++;
        	}
        	
        	else
        	{	
        		ans = (ans*fact(count))%mod;
        		count = 1;
        	}
        }
        
        ans = (ans*fact(count))%mod;
        
       // System.out.println(ans);
        
        boolean ssort = true;
        for(int i = 1; i < n; i++)
        {
        	if(pr[i].b < pr[i-1].b)
        	{
        		ssort = false;
        		break;
        	}
        }
        long sub = 0L;
        if(ssort)
        {
        sub = 1L;
        count = 1;
        for(int i = 1; i < n; i++)
        {	
        	
        	if(pr[i-1].b == pr[i].b && pr[i-1].a == pr[i].a)
        	{
        		count++;
        	}
        	
        	else
        	{	
        		sub = (sub*fact(count))%mod;
        		count = 1;
        	}
        }
        sub = (sub*fact(count))%mod;
        //System.out.println(sub);
        }
        
        count = 1;
        long rans = 1L;
        Arrays.sort(pr,(p,q)->p.b-q.b);
        for(int i = 1; i < n; i++)
        {
        	if(pr[i-1].b == pr[i].b)
        	{
        		count++;
        	}
        	
        	else
        	{	
        		rans = (rans*fact(count))%mod;
        		count = 1;
        	}
        }
        
        rans = (rans*fact(count))%mod;
        ans = (ans+rans)%mod;
        //System.out.println(ans);
        ans = (ans-sub+mod)%mod;
        
        long fa = fact(n);
        //System.out.println(fa);
        fa = (fa-ans+mod)%mod;
        
        out.println(fa);
        out.close();
	}

}
