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.*;
import java.util.*;
public class Main
{
	long fact[];
	void fact(){
		fact=new long[maxn+1];
		fact[0]=1;
		for(int i=1;i<=maxn;i++)
			fact[i]=(fact[i-1]*i)%mod;
	}
	final int maxn=(int)(1e6);
	final long mod=998244353;
	private void solve()throws IOException
	{
		fact();
		int n=nextInt();
		Pair s[]=new Pair[n+1];
		int equala[]=new int[n+1];
		int equalb[]=new int[n+1];
		HashMap<String,Integer> equalboth=new HashMap<>();
		for(int i=1;i<=n;i++)
		{
			int x=nextInt();
			int y=nextInt();
			s[i]=new Pair(x,y);
			equala[x]++;
			equalb[y]++;
			equalboth.put(x+" "+y,equalboth.getOrDefault(x+" "+y,0)+1);
		}
		long curra=1,currb=1,currab=1;
		for(int i=1;i<=n;i++)
		{
			if(equala[i]>1)
				curra=(curra*fact[equala[i]])%mod;
			if(equalb[i]>1)
				currb=(currb*fact[equalb[i]])%mod;
		}
		for(int i:equalboth.values())
			if(i>1)
				currab=(currab*fact[i])%mod;
		boolean bothsorted=true;
		Arrays.sort(s,1,n+1,new Comparator<Pair>(){
			public int compare(Pair a,Pair b){
				return a.first==b.first?a.second-b.second:a.first-b.first;
			}
		});
		for(int i=1;i<n;i++)
		{
			if(s[i].first<=s[i+1].first && s[i].second<=s[i+1].second);
			else
				bothsorted=false;
		}
		if(bothsorted)
			out.println((fact[n]-((curra+currb)%mod-currab+mod)%mod+mod)%mod);
		else
			out.println((fact[n]-(curra+currb)%mod+mod)%mod);
	}
	class Pair{
	    int first,second;
	    Pair(int a,int b){
	        first=a;
	        second=b;
	    }
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws IOException
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();

		br.close();
		out.close();
	}
	public static void main(String args[])throws IOException{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws IOException{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws IOException{
		return br.readLine();
	}
	int nextInt()throws IOException{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws IOException{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws IOException{
		return Double.parseDouble(nextToken());
	}
}