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. 
/******************************************************************************

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*******************************************************************************/
import java.util.*;
public class Main
{
    public static class pair {
        Long a;
        Long b;
        public pair(long x, long y)
        {
            a=x;
            b=y;
        }
    }
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		long md=998244353;
	    pair[] arr=new pair[n];
	    pair[] brr=new pair[n];
	    pair[] crr;
	    long[] fact=new long[400000];
	    fact[0]=1;
	    for(long i=1;i<400000;i++)
	    fact[(int)i]=(i*fact[(int)(i-1)])%md;
	    long ans2=fact[n];
	    for(int i=0;i<n;i++)
	    {
	        long x=sc.nextLong();
	        long y=sc.nextLong();
	        pair p=new pair(x,y);
	        arr[i]=p;
	        brr[i]=p;
	    }
	    Arrays.sort(arr,new Comparator<pair>(){
	       @Override
	       public int compare(pair x,pair y)
	       {
	           return x.a.compareTo(y.a);
	       }
	    });
	    Arrays.sort(brr,(p1,p2) -> p1.b.compareTo(p2.b));
	    crr=Arrays.copyOf(arr,n);
	    long ans=1;
	    for(int i=0;i<n;)
	    {
	        int l=0;
	        long x=arr[i].a;
	        int j=i;
	        while((i<n)&&(arr[i].a==x))
	        {
	            l++;i++;
	        }
	        Arrays.sort(crr,j,i,(p1,p2) -> p1.b.compareTo(p2.b));
	        ans*=fact[l];
	        ans%=md;
	    }
	    long ans1=1;
	    for(int i=0;i<n;)
	    {
	        int l=0;
	        long x=brr[i].b;
	        while((i<n)&&(brr[i].b==x))
	        {
	            l++;i++;
	        }
	        ans1*=fact[l];
	        ans1%=md;
	    }
	    int l=0;long a=crr[0].a,b=crr[0].b;
	    for(int i=1;i<n;i++)
	    {
	        if((crr[i].a>=a)&&(crr[i].b>=b))
	        {
	            a=crr[i].a;
	            b=crr[i].b;
	        }
	        else
	        {
	            l=-1;
	            break;
	        }
	    }
	    long ans3=0;
	    if(l==0)
	    {
	    ans3=1;
	    for(int i=0;i<n;)
	    {
	        int k=0;
	        long x=crr[i].a;
	        long y=crr[i].b;
	        while((i<n)&&(crr[i].a==x)&&(crr[i].b==y))
	        {
	            k++;i++;
	        }
	        ans3*=fact[k];
	        ans3%=md;
	    }
	    }
	    long Ans=ans2;
	    Ans=(Ans-ans1+md)%md;
	    Ans=(Ans-ans+md)%md;
	    Ans=(Ans+ans3)%md;
	    System.out.println(Ans);
	}
}
