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 {
	static BufferedReader br;
	static PrintWriter pr;
	static int cin() throws Exception
	{
		return Integer.valueOf(br.readLine());
	}
	static int[] split() throws Exception
	{
		String[] cmd=br.readLine().split(" ");
		int[] ans=new int[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Integer.valueOf(cmd[i]);
		}
		return ans;
	}
	static long[] splitL() throws IOException
	{
		String[] cmd=br.readLine().split(" ");
		long[] ans=new long[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Long.valueOf(cmd[i]);
		}
		return ans;
	}
	static int sum(long n)
	{
		int ans=0;
		while(n!=0)
		{
			ans=ans+(int)(n%10);
			n=n/10;
		}
		return ans;
	}
	static int gcd(int x,int y)
	{
		if(x==0)
			return y;
		if(y==0)
			return x;
		if(x>y)
			return gcd(x%y,y);
		else
			return gcd(x,y%x);
	}
	static long mod;
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		br=new BufferedReader(new InputStreamReader(System.in));
		pr=new PrintWriter(new OutputStreamWriter(System.out));
		int cases=1;
		long[]fac=new long[300001];
		long res=1;
		fac[0]=1;
		mod=998244353;
		for(int i=1;i<=300000;i++)
		{
		    res=(res*(long)i)%mod;
		    fac[i]=res;
		}
		while(cases!=0)
		{
			cases--;
			int n=cin();
			int[][]arr=new int[n][2];
			for(int i=0;i<n;i++)
			{
			    arr[i]=split();
			}
			long ans=fac[n];
			int[]cnt1=new int[300001];
			int[]cnt2=new int[300001];
			for(int i=0;i<n;i++)
			{
			    cnt1[arr[i][0]]++;
			    cnt2[arr[i][1]]++;
			}
			long x=1;
			long y=1;
			for(int i=1;i<=300000;i++)
			{
			    x=(x*fac[cnt2[i]])%mod;
			}
			for(int i=1;i<=300000;i++)
			{
			    y=(y*fac[cnt1[i]])%mod;
			}
			long z=1;
			Arrays.sort(arr,new Comparator<int[]>(){
			   public int compare(int[]a,int[]b)
			   {
			       if(a[0]!=b[0])
			       return a[0]-b[0];
			       return a[1]-b[1];
			   }
			});
		//	System.out.println(Arrays.deepToString(arr));
			boolean b=true;
			int pr=-1;
			for(int i=0;i<n;i++)
			{
			    if(arr[i][1]<pr)
			    {
			        b=false;
			        break;
			    }
			    pr=arr[i][1];
			}
			if(!b)
			z=0;
			else
			{
			    HashMap<String,Integer>map=new HashMap<>();
			    for(int i=0;i<n;i++)
			    {
			        String f=arr[i][0]+" "+arr[i][1];
			        if(map.containsKey(f))
			        map.put(f,map.get(f)+1);
			        else
			        map.put(f,1);
			    }
			    Iterator<Map.Entry<String,Integer>> iter=map.entrySet().iterator();
			    while(iter.hasNext())
			    {
			        Map.Entry<String,Integer> ele=iter.next();
			        int ct=ele.getValue();
			        z=(z*fac[ct])%mod;
			    }
			}
			//System.out.println(ans+" "+x+" "+y+" "+z);
			ans=((ans-x)%mod+mod)%mod;
			ans=((ans-y)%mod+mod)%mod;
			ans=(ans+z)%mod;
			System.out.println(ans);
		}
	}
}