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.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.Scanner;

public class Main {
	public static void main(String [] args) {
		zapol();
		Scanner sc =new Scanner(System.in);
		int n=sc.nextInt();
		ArrayList<pair> ar=new ArrayList<pair>();
		for (int i=0;i<n;i++) {
			pair w =new pair(sc.nextInt(),sc.nextInt());
			ar.add(w);
		}
		
		Comparator<pair> comp=new Comparator<pair>() {

			@Override
			public int compare(pair o1, pair o2) {
				if (o1.mas[0][0]>o2.mas[0][0]) {
					return 1;
				}
				if (o1.mas[0][0]<o2.mas[0][0]) {
					return -1;
				}
				if (o1.mas[0][1]>o2.mas[0][1]) {
					return 1;
				}
				if (o1.mas[0][1]<o2.mas[0][1]) {
					return -1;
				}
				return 0;
			}
			
		};
		Collections.sort(ar, comp);
		
		long dod=0;
		boolean f=false;
		for (int i=1;i<n;i++) {
			if (ar.get(i).mas[0][1]<ar.get(i-1).mas[0][1]) {
				f=true;
				break;
			}
		}
		if (!f) {
			dod=calc1(ar,0,n);
			dod%=N;
		}
		
		long ans=calc(ar,0,n,0);
		ans%=N;
		
		
	
		 comp=new Comparator<pair>() {

			@Override
			public int compare(pair o1, pair o2) {
				if (o1.mas[0][1]>o2.mas[0][1]) {
					return 1;
				}
				if (o1.mas[0][1]<o2.mas[0][1]) {
					return -1;
				}
				if (o1.mas[0][0]>o2.mas[0][0]) {
					return 1;
				}
				if (o1.mas[0][0]<o2.mas[0][0]) {
					return -1;
				}
				return 0;
			}
			
		};
		Collections.sort(ar, comp);
		long temp=calc(ar,0,n,1);
		temp%=N;
		ans+=temp;
		
		ans-=dod;
		if (ans<0) {
			ans+=N;
		}
		ans%=N;
		ans=dic[n]-ans;
		if (ans<0) {
			ans+=N;
		}
		System.out.println(ans);
		}
	static long calc (ArrayList<pair> ar,int start,int end,int u) {
		int z=1;
		long ans=1;
		for (int i=start+1;i<end;i++) {
			if (ar.get(i).mas[0][u]==ar.get(i-1).mas[0][u]) {
				z++;
			}else {
				ans*=dic[z];
				ans%=N;
				z=1;
			}
		}
		ans*=dic[z];
		ans%=N;
		z=1;
		return ans;
	}
	static long calc1 (ArrayList<pair> ar,int start,int end) {
		int z=1;
		long ans=1;
		for (int i=start+1;i<end;i++) {
			if (ar.get(i).mas[0][0]==ar.get(i-1).mas[0][0]&&ar.get(i).mas[0][1]==ar.get(i-1).mas[0][1]) {
				z++;
			}else {
				ans*=dic[z];
				ans%=N;
				z=1;
			}
		}
		ans*=dic[z];
		ans%=N;
		z=1;
		return ans;
	}
	static long [] dic =new long [1000000];
	static long N=998244353 ;
	static void zapol() {
		long st=1;
		for (int i=1;i<1000000;i++) {
			st*=i;
			st%=N;
			dic[i]=st;
		}
	}
}
class pair{
	public long [][] mas;
	public pair(int x,int y) {
		mas=new long[1][2] ;
		mas[0][0]=x;
		mas[0][1]=y;
	}
}