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.*;
import java.io.*;

public class Solution{

	public static Integer INT(String s){
		return Integer.parseInt(s);
	}

	public static Long LONG(String s){
		return Long.parseLong(s);
	}

	//====================================================================================================================

	static class Pair implements Comparable<Pair>{
		int first, second;
		Pair(int first, int second){
			this.first=first;
			this.second=second;
		}
		public int compareTo(Pair obj){
			if(obj.first==this.first)
				return this.second-obj.second;
			return this.first-obj.first;
		}
	}

	public static int mul(int a, int b){
		return (int)((1L*a*b)%998244353);
	}

	public static int sum(int a, int b){
		return (a+b)%998244353;
	}

	public static void swap(Pair pair){
		int temp=pair.first;
		pair.first=pair.second;
		pair.second=temp;
	}

	public static void main(String args[])throws IOException{

		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	Scanner in=new Scanner(System.in); StringBuilder out=new StringBuilder();

		int n=in.nextInt();

		Pair pair[]=new Pair[n];
		for(int i=0; i<n; i++)
			pair[i]=new Pair(in.nextInt(), in.nextInt());
		
		int fact[]=new int[n+1];
		fact[0]=1;
		for(int i=1; i<=n; i++)
			fact[i]=mul(i, fact[i-1]);

		//-------------------------

		int res=fact[n];

		for(int i=0; i<2; i++){

			Arrays.sort(pair);
			int cnt=1, l=0, r;

			while(l<n){
				r=l+1;
				while(r<n && pair[l].first==pair[r].first) r++;
				cnt=mul(cnt, fact[r-l]);
				l=r;
			}

			res=sum(res, 998244353-cnt);

			for(Pair item: pair)
				swap(item);
		}

		Arrays.sort(pair);
		
		int l=0, r, d=1;
		HashMap<Integer, Integer> map=new HashMap<>();

		while(l<n){
			r=l+1;
			while(r<n && pair[l].first==pair[r].first) r++;
			map.clear();
			for(int i=l; i<r; i++)
				map.put(pair[i].second, map.getOrDefault(pair[i].second, 0)+1);
			for(int item: map.values())
				d=mul(d, fact[item]);
			l=r;
		}

		for(int i=1; i<n; i++)
			if(pair[i-1].second>pair[i].second)
				d=0;

		res=sum(res, d);
		System.out.println(res);
	}
}