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 Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=998244353;
	long gcd(long a,long b) {
		return a==0?b:b>=a?gcd(b%a,a):gcd(b,a);
	}
	void work() {
		int n=ni();
		long[][] A=new long[n][];
		for(int i=0;i<n;i++) {
			A[i]=na(2);
		}
		long[] rec=new long[n+1];
		long ret=1;
		for(int i=1;i<=n;i++) {
			ret=(ret*i)%mod;
			rec[i]=ret;
		}
		long c1=1,c2=1,c3=0;
		Arrays.sort(A,new Comparator<long[]>() {
			public int compare(long[] A1,long[] A2) {
				return (int)(A1[0]-A2[0]);
			}
		});
		for(int i=0,j=0;j<n;j++) {
			long v=A[i][0];
			while(j<n&&A[j][0]==v) {
				j++;
			}
			c1=c1*rec[(j-i)]%mod;
			i=j;
		}
		Arrays.sort(A,new Comparator<long[]>() {
			public int compare(long[] A1,long[] A2) {
				if(A1[0]==A2[0]) {
					return (int)(A1[1]-A2[1]);
				}else {
					return (int)(A1[0]-A2[0]);
				}
			}
		});
		boolean f=true;
		for(int i=1;i<n;i++) {
			if(A[i][0]<A[i-1][0]||A[i][1]<A[i-1][1]) {
				f=false;
				break;
			}
		}
		if(f) {
			c3=1;
			for(int i=0,j=0;j<n;j++) {
				long v1=A[i][0],v2=A[i][1];
				while(j<n&&A[j][0]==v1&&A[j][1]==v2) {
					j++;
				}
				c3=c3*rec[(j-i)]%mod;
				i=j;
			}
		}
			
		Arrays.sort(A,new Comparator<long[]>() {
			public int compare(long[] A1,long[] A2) {
				return (int)(A1[1]-A2[1]);
			}
		});
		
		for(int i=0,j=0;j<n;j++) {
			long v=A[i][1];
			while(j<n&&A[j][1]==v) {
				j++;
			}
			c2=c2*rec[(j-i)]%mod;
			i=j;
		}
		ret=((ret-c1-c2+c3)%mod+mod)%mod;
		out.println(ret);
	}




	//input
	private ArrayList<Integer>[] ng(int n, int m) {
		ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			int s=in.nextInt()-1,e=in.nextInt()-1;
			graph[s].add(e);
			graph[e].add(s);
		}
		return graph;
	}

	private int ni() {
		return in.nextInt();
	}

	private long nl() {
		return in.nextLong();
	}

	private long[] na(int n) {
		long[] A=new long[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextLong();
		}
		return A;
	}
}	

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}