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.*;
public class CF597D{
    static class Pair{
        int x,y;
        Pair(int x,int y){
            this.x=x;
            this.y=y;
        }
    }
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt(),i;
        long m=998244353;
        ArrayList<Pair> arr=new ArrayList<>();
        for(i=0;i<n;i++){
            arr.add(new Pair(sc.nextInt(),sc.nextInt()));
        }
        Collections.sort(arr,new Comparator<Pair>(){
            public int compare(Pair p1,Pair p2){
                if(p1.x==p2.x) return 0;
                return p1.x>p2.x?1:-1;
            }
        });
        long fact[]=new long[300001],sorted=1,a;
        fact[0]=1;
        for(i=1;i<=300000;i++){
            fact[i]=(fact[i-1]*i)%m;
        }
        int l=0,r=0;
        while(l<n){
            r=l+1;
            while(r<n&&arr.get(r).x==arr.get(l).x) r++;
            sorted=(sorted*fact[r-l])%m;
            l=r;
        }
        a=sorted;
        //System.out.println(a);
        sorted=1;
        l=0;r=0;
        Collections.sort(arr,new Comparator<Pair>(){
            public int compare(Pair p1,Pair p2){
                if(p1.y==p2.y) return 0;
                return p1.y>p2.y?1:-1;
            }
        });
		/*for(Pair p:arr){
		    System.out.println(p.x+" "+p.y);
		}*/
        while(l<n){
            r=l+1;
            while(r<n&&arr.get(r).y==arr.get(l).y) r++;
            //System.out.println("Index: "+r);
            sorted=(sorted*fact[r-l])%m;
            l=r;
        }
        a=(a+sorted)%m;
        //System.out.println(sorted);
        l=0;r=0;
        long d=1;
        while(l<n){
            r=l+1;
            while(r<n&&(arr.get(l).x==arr.get(r).x&&arr.get(l).y==arr.get(r).y)){
                r++;
            }
            //System.out.println("Index: "+r);
            d=(d*fact[r-l])%m;
            l=r;
        }
        //System.out.println(d);
        for(i=1;i<arr.size();i++){
            if(arr.get(i-1).x>arr.get(i).x) d=0;
        }
        long res=(fact[n]-a+m)%m;
        res=(res+d)%m;
        System.out.println(res);
    }
}